Calculate value of $\displaystyle x$ in $\displaystyle (7x+1)^{\frac{1}{3}}+(-x^2+x+8)^{\frac{1}{3}}+(x^2-8x-1)^{\frac{1}{3}}=2$
How can I say.....
Here is my process...
Let $\displaystyle (7x+1)^{\frac{1}{3}}=a$
$\displaystyle (-x^2+x+8)^{\frac{1}{3}}=b$
$\displaystyle (x^2-8x-1)^{\frac{1}{3}}=c$
So We Get $\displaystyle a+b+c=a^3+b^3+c^3$
$\displaystyle a(a^2-1)+b(b^2-1)+c(c^2-1)=0$
Now I am struck at that point.......