Question

If $x^a\cdot x^b\cdot x^c=1$, then $a^3+b^3+c^3$ is equal to:

• A. $9$
• B. $abc$
• C. $a+b+c$
• D. $3abc$

Answer 1

The correct option is D $3abc$

Given,

$x^a\cdot x^b\cdot x^c=1$.

Then, on simplifying,

$x^a\cdot x^b\cdot x^c=1$

$⟹$ $(x{)}^{a+b+c}=1$

$⟹$ $(x{)}^{a+b+c}=(x{)}^0$

$⟹$ $a+b+c=0---\left(i\right)$.

We know,

$(a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Substitute $\left(i\right)$, we get,

$(a^3+b^3+c^3)-3abc=\left(0\right)(a^2+b^2+c^2-ab-bc-ca)$

$⟹$$(a^3+b^3+c^3)-3abc=0$

$⟹$ $a^3+b^3+c^3=3abc$.

Hence, option $D$ is correct.