Algebra (Thailand Math POSN 2nd round)

Write a full solution,

  1. Find all roots of x5+x4+x3x2x1=0x^{5}+x^{4}+x^{3}-x^{2}-x-1 = 0.

  2. Prove that (1+sinθ+icosθ1+sinθicosθ)n=cos(nπ2nθ)+isin(nπ2nθ)\displaystyle \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)

  3. If a,b,c,da,b,c,d are roots of equation x412x3+54x2118x+96=0x^{4}-12x^{3}+54x^{2}-118x+96 = 0, then find the value of (a3)4+(b3)4+(c3)4+(d3)4(a-3)^{4}+(b-3)^{4}+(c-3)^{4}+(d-3)^{4}.

  4. Let P(x),Q(x)P(x),Q(x) be polynomial with real coefficients such that deg(P(x))>deg(Q(x))\deg(P(x)) > \deg(Q(x)). Find all solutions (P(x),Q(x))(P(x),Q(x)) that satisfy P(x)2+Q(x)2=x6+1P(x)^{2}+Q(x)^{2} = x^{6}+1.

This note is a part of Thailand Math POSN 2nd round 2015

Note by Samuraiwarm Tsunayoshi
4 years, 7 months ago

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Solution for Q-3:

Using binomial theorem, one can note that,

P(x)=(x3)410x+15P(x)=(x-3)^4-10x+15

We have, using Remainder-Factor Theorem and the fact that a,b,c,da,b,c,d are roots of P(x)P(x) that,

P(x)=(x3)410x+15=0  x{a,b,c,d}    (x3)4=10x15  x{a,b,c,d}P(x)=(x-3)^4-10x+15=0~\forall~x\in\{a,b,c,d\}\implies (x-3)^4=10x-15~\forall~x\in\{a,b,c,d\}

Using the last result, we have the sum as,

x{a,b,c,d}(x3)4=x{a,b,c,d}(10x15)\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)

Using Vieta's formulas, we have,

x{a,b,c,d}(x)=12    x{a,b,c,d}(10x)=120\displaystyle \sum_{x\in\{a,b,c,d\}} (x) = 12\\ \implies \sum_{x\in\{a,b,c,d\}} (10x)=120

Hence, the required sum evaluates as,

x{a,b,c,d}(x3)4=x{a,b,c,d}(10x15)={(x{a,b,c,d}(10x))(x{a,b,c,d}(15))}=12015×4=12060=60\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)=\left\{\left(\sum_{x\in\{a,b,c,d\}} (10x)\right)-\left(\sum_{x\in\{a,b,c,d\}} (15)\right)\right\}=120-15\times 4=120-60=\boxed{60}

Prasun Biswas - 4 years, 6 months ago

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Nicely done.

Calvin Lin Staff - 4 years, 6 months ago

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Your hint did most of the work. So, the credit actually goes to you. :)

Prasun Biswas - 4 years, 6 months ago

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Q1Q-1

x5+x4+x3x2x1=(x1)(x4+2x3+3x2+2x+10)x^5+x^4+x^3-x^2-x-1 = (x-1)(x^4+2x^3+3x^2+2x+10)    one root is 1\implies \text{one root is 1}

x4+2x3+3x2+2x+1=0    x2+2x+3+2x+1x2=0x^4+2x^3+3x^2+2x+1=0\implies x^2+2x +3+\dfrac{2}{x}+\dfrac{1}{x^2}=0     (x+1x)22+2(x+1x)+3=0\implies \left(x+\dfrac{1}{x}\right)^2-2+2\left(x+\dfrac{1}{x}\right)+3=0

x+1x=mx+\dfrac{1}{x} = m

m2+2m+1=0    (m+1)2=0    m=1,1m^2+2m+1=0 \implies \left(m+1\right)^2=0\implies m=1,1

x+1x=1On solving using quadratic equations we get x=ω,ω2x+\dfrac{1}{x}=1 \text{On solving using quadratic equations we get x}=\omega,\omega^2

(    we get 4 roots from here ω,ω,ω2,ω2\implies \text{we get 4 roots from here }\omega,\omega,\omega^2,\omega^2)

roots are     \implies ω,ω2,1,ω,ω2\boxed{\omega,\omega^2,1,\omega,\omega^2}

Parth Lohomi - 4 years, 7 months ago

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You can factor it the easier way: x5+x4+x3x2x1=x3(x2+x2+1)(x2+x+1)=(x31)(x2+x+1)x^5 + x^4 + x^3 - x^2 - x - 1 = x^3(x^2 + x^2 + 1) - (x^2 + x + 1) = (x^3-1)(x^2+x+1)

(x1)(x2+x+1)2=0x=1,ω,ω2 \Rightarrow (x-1)(x^2+x+1)^2 = 0 \Rightarrow x = 1, \omega, \omega^2

Pi Han Goh - 4 years, 7 months ago

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What motivated you to factor out x2+x+1x^{2}+x+1 ?

Aditya Sky - 3 years, 5 months ago

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@Aditya Sky Looking at Parth's solution tells us that x2+x+1x^2 + x+ 1 is a factor, so I just factor x2+x+1x^2+x+1 from the start.

Pi Han Goh - 3 years, 5 months ago

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4) deg(P2)=2deg(P)>2degQ=deg(Q2) \deg(P^2) = 2 \deg(P) > 2 \deg Q = \deg(Q^2) .

6=deg(x6+1)=deg(P2+Q2)=deg(P2)=2deg(P)6 = \deg(x^6 + 1) = \deg(P^2 + Q^2) = \deg(P^2) = 2\deg(P) .

deg(P)=3deg(Q)2deg(Q2)4P2=1x6+0x5+\deg(P) = 3 \Rightarrow \deg(Q) \leq 2 \Rightarrow \deg(Q^2) \leq 4 \Rightarrow P^2 = 1x^6 +0x^5 + \cdots .

This means P=ux3+p1x+p0 P = ux^3 + p_1x + p_0 and Q=q2x2+q1x+q0Q = q_2 x^2 + q_1x + q_0 where p1,p0,q2,q1,q0Rp_1, p_0, q_2, q_1, q_0 \in \mathbb R and u=±1u = \pm 1.

From P2+Q2=x6+1P^2 + Q^2 = x^6 + 1 we get the following equations

{p02+q02=1 p0p1+q0q1=0 p12+q12+2q0q2=0 up0+q1q2=0 2up1+q22=0 \displaystyle{ \cases{ p_0^2 + q_0^2 = 1 \\ \ \\ p_0p_1 + q_0q_1 = 0 \\ \ \\ p_1^2 + q_1^2 +2q_0q_2 = 0\\ \ \\ up_0 + q_1q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0 } }

The first one tell us there is α[0,2π)\alpha \in [0, 2\pi) such that

(p0,q0)=(cosα,sinα) (p_0, q_0) = (\cos \alpha, \sin \alpha)

The second equation states that

(p0,q0)(p1,q1) (p_0, q_0) \perp (p_1, q_1)

so must exist ρR \rho \in \mathbb R such that

(p1,q1)=(ρsinα,ρcosα) (p_1, q_1) = (-\rho \sin \alpha, \rho \cos \alpha)

Assuming cosα0 \cos\alpha \not= 0 the fourth equation becomes

q2=uρ q_2 = -\frac{u}{\rho}

Assuming sinα0 \sin\alpha \not= 0 the fifth and the third equations become

q22=2uρsinαq2=ρ22sinα q_2^2 = 2u\rho\sin\alpha \\ q_2 = - \frac{\rho^2}{2\sin\alpha}

Squaring the latter we get

ρ44sin2α=q22=2uρsinα \frac{\rho^4}{4\sin^2\alpha} = q_2^2 = 2u\rho\sin\alpha

and so

ρ3=8usin3α \rho^3 = 8u\sin^3\alpha

By the iniectivity of the function f(x)=x3f(x) = x^3 and by the obvious fact that u3=uu^3 = u we get

ρ=2usinα \rho = 2u\sin\alpha

Using this equality into

q2=uρ and q2=ρ22sinα q_2 = -\dfrac{u}{\rho} \quad \textrm{ and } \quad q_2 = - \dfrac{\rho^2}{2\sin\alpha}

we see that

q2=2sinα and sin2α=14 q_2 = -2\sin\alpha \quad \textrm{ and } \quad \sin^2\alpha = \dfrac{1}{4}

So we find sinα=±12\sin\alpha = \pm \dfrac{1}{2} and the possible values for α \alpha are

α=π6,5π6,7π6,11π6\alpha = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}

and the following identities

{p0=cosα q0=sinα p1=2usin2α=2uqo2 q1=2usinαcosα=2uq0p0 q2=2sinα=2q0 \displaystyle{ \cases{ p_0 = \cos\alpha \\ \ \\ q_0 = \sin\alpha \\ \ \\ p_1 = -2u\sin^2\alpha = -2uq_o^2\\ \ \\ q_1 = 2u\sin\alpha\cos\alpha = 2uq_0p_0 \\ \ \\ q_2 = -2\sin\alpha = -2q_0 } }

Putting in the possible values for α\alpha we get the following 8 couples of polynomials (each vale giving two polynomials depending on uu being 11 or 1-1 ) that we can write in short in the following way.

Define

A=x3x2+32 and B=x232x12 A =x^3 - \dfrac{x}{2} + \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad B = x^2 - \dfrac{\sqrt3}{2}x - \dfrac{1}{2} C=x3x232 and D=x2+32x12 C =x^3 - \dfrac{x}{2} - \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad D = x^2 + \dfrac{\sqrt3}{2}x - \dfrac{1}{2}

then the following are solution of our problem

(P,Q)={(±A,±B) (±C,±D) (P,Q) =\cases{ (\pm A , \pm B) \\ \ \\ (\pm C , \pm D) }

This solution left out the cases in which cosαsinα=0\cos \alpha \cdot \sin \alpha = 0 . These cases lead to trivial solutions as we can easily find out.

If cosα=0\cos\alpha = 0 then p0=0p_0 = 0 and q0=v=±1q_0 = v = \pm1, and we immediately get q1=0q_1 = 0 and p1=vρp_1 = -v\rho. The equations {p12+q12+2q0q2=0 2up1+q22=0become{ρ2+2vq2=0 2uvρ+q22=0\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 +2vq_2 = 0 \\ \ \\ -2uv\rho + q_2^2 = 0 }

{ρ(ρ38uv)=0 q2=ρ22v\cases{\rho(\rho^3 -8uv) = 0 \\ \ \\ q_2= - \dfrac{\rho^2}{2v} }

By the upper equation we have either ρ=0\rho = 0 or ρ=2uv\rho = 2uv If ρ=0\rho = 0 we get q2=p1=0q_2=p_1=0 so in this case

{p0=0 q0=v=±1 p1=0 q1=0 q2=0 \displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = 0\\ \ \\ q_1 = 0 \\ \ \\ q_2 = 0 } }

that leads to these four couple of solutions

(P,Q)=(±x3,±1) (P,Q) = (\pm x^3 , \pm 1)

If ρ=2uv\rho = 2uv we get q2=2v,p1=2uq_2= -2v, p_1= -2u so in this case

{p0=0 q0=v=±1 p1=2u q1=0 q2=2v \displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = -2u\\ \ \\ q_1 = 0 \\ \ \\ q_2 = -2v } }

that gives these other four couples

(P,Q)=(±(x32x),±(2x21)) (P,Q) = (\pm (x^3 -2x), \pm (2x^2 -1) )

Last case is when sinα=0 \sin \alpha = 0 . In this case we get q0=p1=0,p0=v,q1=ρvq_0 = p_1 = 0, p_0 = v, q_1 = \rho v where v=±1v = \pm 1 . The equations

{p12+q12+2q0q2=0 2up1+q22=0become{ρ2=0 q22=0\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 = 0 \\ \ \\ q_2^2 = 0 }

so ρ=q2=q1=0\rho=q_2=q_1=0 and the equation up0+q1q2=0up_0 + q_1q_2 = 0 becomes

uv=0uv = 0

that is always false. So this latter case doesn't lead to any further solution and our analysis is complete.

We found 16 pairs of polynomials (P,Q)(R[x])2 (P, Q) \in (\mathbb R[x])^2 that satisfy the equation.

Andrea Palma - 4 years, 6 months ago

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It's not clear to me what you're trying to do here. Are you saying that no such polynomials exist?

Calvin Lin Staff - 4 years, 6 months ago

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I don't know! I ended up finding 8 couples of polynomials. I was sure I posted the solution but I guess my internet connection (or me) failed to post the whole solution. I will repost it in short time. I'm writing right now.

-----EDIT---- Just wrote the solution that I made this morning. I just realized that I didn't examine some cases, and so my solution is not complete, but the cases left put are easy to handle (I feel) and I'll do as soon as possible.

--- EDIT --- Added also the trivial cases (boring). Now I think the problem is completely discussed and solved.

Andrea Palma - 4 years, 6 months ago

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A more straightforward solution for Q-2 using Calvin's hint:

We first modify the expression of LHS by multiplying numerator and denominator of the expression inside brackets by (1+sinθ+icosθ)(1+\sin\theta+i\cos\theta).

LHS=(1+sinθ+icosθ1+sinθicosθ)n=((1+sinθ+icosθ)2(1+sinθ)2(icosθ)2)n    LHS=(1cos2θ+sin2θ+2sinθ+2icosθ+2icosθsinθ1+sin2θ+cos2θ+2sinθ)n    LHS=(2sinθ+2sin2θ+2icosθ+2icosθsinθ2+2sinθ)n    LHS=(2(1+sinθ)(sinθ+icosθ)2(1+sinθ))n=(sinθ+icosθ)n    LHS=(cos(π2θ)+isin(π2θ))n\textrm{LHS}=\left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^n=\left(\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2-(i\cos\theta)^2}\right)^n\\ \implies \textrm{LHS}=\left(\frac{1-\cos^2\theta+\sin^2\theta+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta}{1+\sin^2\theta+\cos^2\theta+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2\sin\theta+2\sin^2\theta+2i\cos\theta+2i\cos\theta\sin\theta}{2+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2(1+\sin\theta)(\sin\theta+i\cos\theta)}{2(1+\sin\theta)}\right)^n=(\sin\theta+i\cos\theta)^n\\ \implies \textrm{LHS}=\bigg(\cos\left(\frac{\pi}{2}-\theta\right)+i\sin\left(\frac{\pi}{2}-\theta\right)\bigg)^n

Using Euler's formula eiθ=cosθ+isinθe^{i\theta}=\cos\theta+i\sin\theta and laws of indices, we can modify LHS as,

LHS=(e(π2θ))n=en(π2θ)=e(nπ2nθ)=cos(nπ2nθ)+isin(nπ2nθ)=RHS\textrm{LHS}=\large \left(e^{\left(\frac{\pi}{2}-\theta\right)}\right)^n=e^{n\left(\frac{\pi}{2}-\theta\right)}=e^{\left(\frac{n\pi}{2}-n\theta\right)}=\cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)=\textrm{RHS}

(1+sinθ+icosθ1+sinθicosθ)n=cos(nπ2nθ)+isin(nπ2nθ)\therefore\quad \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)

And we are done. _\square


Elementary identities used:

  • cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta=1
  • (a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
  • cos(π2θ)=sinθ\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta
  • sin(π2θ)=cosθ\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta

Prasun Biswas - 4 years, 6 months ago

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I also noticed that one can directly get the RHS form without going into Euler's formula by directly using De Moivre's Theorem.

Prasun Biswas - 4 years, 6 months ago

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Q3Q-3

This is not a solution

I Used Newtons sums to get answer 60\boxed{60}

Parth Lohomi - 4 years, 7 months ago

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Did you mean Q3?

Samuraiwarm Tsunayoshi - 4 years, 7 months ago

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Yes.

Parth Lohomi - 4 years, 7 months ago

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There is a pretty simple solution.

Hint: What is (x3)4 (x-3)^4 ?

Calvin Lin Staff - 4 years, 7 months ago

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Yes sir , on opening all the terms I got

P412P3+54P2108P1+324P_4-12P_3+54P_2-108P_1+324

Where Pn=an+bn+cn+dnP_n=a^n+b^n+c^n+d^n

now we can apply newtons sums.

Parth Lohomi - 4 years, 6 months ago

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@Parth Lohomi See Prasun's solution at the top for the "two-line" solution to this problem.

Calvin Lin Staff - 4 years, 6 months ago

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That was quite a big hint. :D

Prasun Biswas - 4 years, 6 months ago

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2) is pretty simple. First, replace θ=π/2a \theta = \pi/2 - a .

Then, 1+sinθ+icosθ=1+cosa+isina=2cos2a/2+i2sin(a/2)cos(a/2)=2cos(a/2)(cosa/2+isina/2) 1 +\sin \theta + i \cos \theta = 1 + \cos a + i \sin a = 2\cos^2a/2 + i2\sin(a/2) \cos(a/2) = 2\cos(a/2)(\cos a/2+i\sin a/2)

Put b=a/2 b = a/2 Likewise, 1+sinθicosθ=2cosb(cosbisinb). 1 + \sin\theta - i\cos \theta = 2\cos b(\cos b-i\sin b).

Then 1+sinθ+icosθ1+sinθicosθ=2cosb(cosb+isinb)2cosb(cosbisinb)=cosb+isinbcosbisinb \dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta } = \dfrac{2\cos b(\cos b+i\sin b)}{2\cos b(\cos b-i\sin b)} = \dfrac{\cos b+i\sin b}{\cos b-i\sin b}

=(cosb+isinb)(cosbisinb)(cosb+isinb)(cosb+isinb)=(cosb+isinb)2cos2b+sin2b=(cosb+isinb)2 = \dfrac{(\cos b+i\sin b)}{(\cos b-i\sin b)} * \dfrac{(\cos b+i\sin b)}{(\cos b+i\sin b)} = \dfrac{(\cos b+i\sin b)^2}{\cos^2 b + \sin^2 b} = (\cos b+i\sin b)^2

Therefore, (1+sinθ+icosθ1+sinθicosθ)n=((cosb+isinb)2)n=(cosb+isinb)2n (\dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta })^n = ((\cos b+i\sin b)^2)^n = (\cos b + i\sin b)^{2n}

=cos2bn+isin2bn=cosan+isinan=cos(n(π/2a))+isin(n(π/2a)) = \cos 2bn + i\sin 2bn = \cos an + i\sin an = \cos(n(\pi/2 - a)) + i\sin(n(\pi/2 - a))

Siddhartha Srivastava - 4 years, 7 months ago

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There's a more straightforward way to solve 2.

Hint: (eiθ)n=eiθn \left( e ^ { i \theta } \right) ^n = e^{i \theta n } .

Calvin Lin Staff - 4 years, 6 months ago

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That was a big hint too. :P

Prasun Biswas - 4 years, 6 months ago

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I'm confused? Isn't Demoivre's theroem equivalent to that fact?

Siddhartha Srivastava - 4 years, 6 months ago

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@Siddhartha Srivastava Yes it is. The hints that I give are often to suggest alternative approaches, that you would have to work out how to apply them. In this case, see Prasun's solution above.

Calvin Lin Staff - 4 years, 6 months ago

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@Calvin Lin I was asking because I had used Demoivre's theorem. I'm also not sure how Prasun's solution is vastly different from mine, since we use almost the same approach. The only major difference in our solutions is the way we simplify the original expression

Siddhartha Srivastava - 4 years, 6 months ago

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@Siddhartha Srivastava Ah, the difference is more in terms of motivation. IE, to explain that the intermediate step is (cos(π2θ)+sin(π2θ))n \left( \cos ( \frac{\pi}{2} - \theta) + \sin ( \frac{\pi}{2} - \theta ) \right) ^ n , which would explain why we did the calculations that we did.

Calvin Lin Staff - 4 years, 6 months ago

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3) a,b,c,d are roots of equation , so

(a-3)^4 = 10a-15 ------m

(b-3)^4 = 10b-15 -------n

(c-3)^4 = 10c-15 -------o

(d-3)^4 = 10d-15 -------p

from vieta's formula we'll get a+b+c+d = 12

m+n+o+p ; (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 10(a+b+c+d)-60 = 10(12)-60 = 60

so, (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 60

Nut Nutthapong - 4 years, 6 months ago

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1.x5+x4+x3x2x1=x5+x4+x3(x2+x+1)=x3(x2+x+1)1(x2+x+1)=(x2+x+1)(x31)=(x2+x+1)(x2+x+1)(x1)Solvingquadratics,wegetroots1,ω,ω,ω2,ω21.\quad { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-{ x }^{ 2 }-x-1\\ ={ x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-\left( { x }^{ 2 }+x+1 \right) \\ ={ x }^{ 3 }\left( { x }^{ 2 }+x+1 \right) -1\left( { x }^{ 2 }+x+1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 3 }-1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 2 }+x+1 \right) \left( x-1 \right) \\ \\ Solving\quad quadratics,\quad we\quad get\quad roots\quad 1,\omega ,\omega ,{ \omega }^{ 2 },{ \omega }^{ 2 }

Archit Boobna - 4 years, 6 months ago

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