Question

If $a+b+c=0,$ then $(a^3+b^3+c^3{)}^2=$...........

• A. $3a^2{b}^2{c}^2$
• B. 9abc
• C. $9a^2{b}^2{c}^2$
• D. 27abc

Answer 1

The correct option is C $9a^2{b}^2{c}^2$

If $a+b+c=0$, then, $a+b=-c....\left(1\right)$

Now, taking cubes on both the sides, we get:

$(a+b{)}^3=(-c{)}^3\Rightarrow a^3+b^3+3ab(a+b)=-c^3(\because {(x+y)}^3=x^3+y^3+3xy(x+y\left)\right)\Rightarrow a^3+b^3+3ab(-c)=-c^3\left(Fromeqn\right(1\left)\right)\Rightarrow a^3+b^3-3abc=-c^3\Rightarrow a^3+b^3+c^3=3abc$

Therefore, $a^3+b^3+c^3\ne 0$ unless $a,b,c=0$

Thus we have, $a^3+b^3+c^3=3abc$

Squaring both sides we get,

${(a^3+b^3+c^3)}^2={\left(3abc\right)}^2\Rightarrow {(a^3+b^3+c^3)}^2=9a^2{b}^2{c}^2$

Hence, ${(a^3+b^3+c^3)}^2=9a^2{b}^2{c}^2$