Tackling Algebra

I AM UNABLE TO TACKLE THIS PROBLEM. CAN ANYONE COME FORWARD TO HELP ME.

a+b+c=-1/2,ab+bc+ca=-1/2,abc=1/8, then how to get 1) 1/a+1/b+1/c and 2)a^1/3+b^1/3+c^1/3.

Note by Prabir Chaudhuri
4 years, 9 months ago

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Let a13=x,b13=y,c13=za^{\frac{1}{3}} = x, b^{\frac{1}{3}} = y ,c^{\frac{1}{3}} = z

then we have

{x3+y3+z3=12(xy)3+(yz)3+(xz)3=12(xyz)3=18\begin{cases} x^3+y^3+z^3=-\dfrac{1}{2}\\ (xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\ (xyz)^3=\dfrac{1}{8} \end{cases}

using the identity

A3+B3+C3=(A+B+C)33(A+B+C)(AB+BC+AC)+3ABCA^3+B^3+C^3=(A+B+C)^3-3(A+B+C)(AB+BC+AC)+3ABC

we have,

{(x+y+z)33(x+y+z)(xy+yz+xz)+3xyz=12(xy+yz+xz)33(xy+yz+xz)[xyz(x+y+z)]+3x2y2z2=12xyz=12\begin{cases} (x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz=-\dfrac{1}{2}\\ (xy+yz+xz)^3-3(xy+yz+xz)[xyz(x+y+z)]+3x^2y^2z^2=-\dfrac{1}{2}\\ xyz=\dfrac{1}{2} \end{cases}

Now,

Let u=x+y+z,v=xy+yz+xzu=x+y+z, v=xy+yz+xz

then we have

{u33uv+2=04v36uv+5=0\begin{cases} u^3-3uv+2=0\\ 4v^3-6uv+5=0 \end{cases}

4v32u3+1=0,v=u3+23u\Longrightarrow 4v^3-2u^3+1=0, v=\dfrac{u^3+2}{3u}

so

4(u3+23u)32u3+1=04u930u6+75u3+32=04\left(\dfrac{u^3+2}{3u}\right)^3-2u^3+1=0\Longrightarrow 4u^9-30u^6+75u^3+32=0

let t=u3t=u^3, so we have

4t330t2+75t+32=04t^3-30t^2+75t+32=0

let t=52at=\dfrac{5}{2}-a,then

4(52a)330(52a)2+75(52a)+32=04\left(\dfrac{5}{2}-a\right)^3-30\left(\dfrac{5}{2}-a\right)^2+75\left(\dfrac{5}{2}-a\right)+32=0

4a3=1892a=3732\Longrightarrow 4a^3=\dfrac{189}{2}\Longrightarrow a=\dfrac{3\sqrt[3]{7}}{2}

t=523732\Longrightarrow t=\dfrac{5}{2}-\dfrac{3\sqrt[3]{7}}{2}

Therefore,

a13+b13+c13=x+y+z=u=t13=12(5373)3a^{\frac{1}{3}} + b^{\frac{1}{3}} + c^{\frac{1}{3}} = x+y+z = u = t^{\frac{1}{3}} = \sqrt[3]{\dfrac{1}{2}\left(5-3\sqrt[3]{7}\right)}

Ishan Singh - 4 years, 8 months ago

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@PRABIR CHAUDHURI I have posted my solution of the other part. You may look at it.

Ishan Singh - 4 years, 8 months ago

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See.. the first part is easy as ab+bc+ca=-1/2divide the equation by abc which is also equal to 1/8 . We get 1/a +1/b +1/c=1/2 * 8 =4.

Mayur Ade - 4 years, 9 months ago

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Thanks.

Prabir Chaudhuri - 4 years, 9 months ago

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1/a+1/b+1/c=(ab+bc+ca)/abc

A Former Brilliant Member - 3 years, 12 months ago

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