21i think my answer is complete
because
a= -b implies a3 = -b3 (f(x) = x^3 is monotonic)
which is extremely easy to solve for
$\textbf{Calculate value of $\mathbf{x}$ in $}\\\\ \mathbf{(7x+1)^{\frac{1}{3}}+(-x^2+x+8)^{\frac{1}{3}}+(x^2-8x-1)^{\frac{1}{3}}=2}$
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8 Answers
21Let (7x+1)1/3 = a
(-x2+x + 8)1/3= b
(x2-8x -1)1/3 = c
observe that a3+b3+c3 = 8
and we have to solve a+b+c = 2
≡ a3+b3+c3 + 3(a+b)(b+c)(a+c) = 8
which means 3(a+b)(b+c)(a+c) = 0
Which is supported by a=-b or b=-c or a=-c
finished,no ?
62(a+b+c)^3=a^3+b^3+c^3
Edits below after shubhodip pointed out..
(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc
So, (a+b+c)(ab+bc+ca)=abc
Which are the roots of the polynomial
t^3-pt^2+qt-pq
=(t^2+q)(t-p)
which means t=p
which means the root is equal to the sum of roots..
so a+b+c = a or b or c
which will lead to a=-b or a=-c or b=-c
Now we are led to one of subhodip's conclusion in the posts below
21i have got roots as
x= 0,1,-1,9
btw, i can't find why wolfram is not able to solve it
2162yup it is ..
getting the same result in the proof that i edited after you pointed out [1]
