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.

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestLet \[a^{\frac{1}{3}} = x, b^{\frac{1}{3}} = y ,c^{\frac{1}{3}} = z\]

then we have

\[\begin{cases} x^3+y^3+z^3=-\dfrac{1}{2}\\ (xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\ (xyz)^3=\dfrac{1}{8} \end{cases}\]

using the identity

\[A^3+B^3+C^3=(A+B+C)^3-3(A+B+C)(AB+BC+AC)+3ABC\]

we have,

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

Now,

Let \[u=x+y+z, v=xy+yz+xz\]

then we have

\[\begin{cases} u^3-3uv+2=0\\ 4v^3-6uv+5=0 \end{cases}\]

\[\Longrightarrow 4v^3-2u^3+1=0, v=\dfrac{u^3+2}{3u}\]

so

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

let \(t=u^3\), so we have

\[4t^3-30t^2+75t+32=0\]

let \(t=\dfrac{5}{2}-a\),then

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

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

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

Therefore,

\[a^{\frac{1}{3}} + b^{\frac{1}{3}} + c^{\frac{1}{3}} = x+y+z = u = t^{\frac{1}{3}} = \sqrt[3]{\dfrac{1}{2}\left(5-3\sqrt[3]{7}\right)} \] – Ishan Singh · 2 years, 1 month ago

Log in to reply

@PRABIR CHAUDHURI I have posted my solution of the other part. You may look at it. – Ishan Singh · 2 years, 1 month ago

Log in to reply

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 · 2 years, 2 months ago

Log in to reply

– Prabir Chaudhuri · 2 years, 2 months ago

Thanks.Log in to reply

Log in to reply

– Siddhartha Srivastava · 2 years, 2 months ago

I think its \( a^{\frac{1}{3}} \) , not \( a^{-3} \).Log in to reply

– Megh Choksi · 2 years, 2 months ago

Oh yes sorry its totally wrong , it was 12:30 nearly I was was half sleep , and I wrote it sorryLog in to reply

– Prabir Chaudhuri · 2 years, 2 months ago

Please note that it is a^(1/3),not a^(-3) as anticipated. Is there any general formulation for evaluating a^(1/n) +b^(1/n) + c^(1/n) with the input as given with the problem ?Thanks.Log in to reply

– Prabir Chaudhuri · 2 years, 2 months ago

Yeah, you are right. It is a^(1/3),not a^(-3).Log in to reply

1/a+1/b+1/c=(ab+bc+ca)/abc – Sanchayan Dutta · 1 year, 5 months ago

Log in to reply