Question

If $\sqrt[3]{3(1-x^3)}(y-z)+\sqrt[3]{3(1-y^3)}(z-x)+\sqrt[3]{3(1-z^3)}(x-y)=0$
then prove that $(1-x{)}^3(1-y{)}^3(1-z{)}^3=(1-xyz{)}^3$

Answer 1

It is a question base on the formula that if A + B + C =
0 then $A^3+B^3+C^3=3ABC$
Let $(1-x^3{)}^{1/3}=aetc.$ y - z = l etc.
We are given that
al + bm + cn = 0 ....(1)
l + m + n = $\sum$ (y - z) = 0 ....(2)
lx + my + nz = $\sum$ x(y - z) = 0 ....(3)
$a^3{l}^3+b^3{m}^3+c^3{n}^3=abclmn,by\left(1\right)$ ....(4)
$x^3{l}^3+y^3{m}^3+z^3{n}^3=xyzlmn,by\left(3\right)$ ....(5)
$l^3+y^3+z^3=3lmn,by\left(2\right)$ .....(6)
Subtracting the last two, we get
${\sum }^3{l}^3(1-x^3)=3lmn(1-xyz)$
$a^3{l}^3+b^3{m}^3+c^3{n}^3=3lmn(1-xyz)\because 1-x^3=a^3$
or 3abclmn = 3lmn (1 - xyz) by (4)
Cancel 3lmn and cube both sides
$\therefore a^3{b}^3{c}^3=(1-xyz{)}^3$
or $(1-x^3)(1-y^3)(1-z^3)=(1-xyz{)}^3$

Hence proved.