# 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
6 years, 1 month ago

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Let $a^{\frac{1}{3}} = x, b^{\frac{1}{3}} = y ,c^{\frac{1}{3}} = z$

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,

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

- 6 years ago

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

- 6 years ago

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.

- 6 years, 1 month ago

Thanks.

- 6 years, 1 month ago

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

- 5 years, 4 months ago