Question

State true or false:

If $a,b\in +R$ and $a\ne b$ then, $a^3+b^3<a^{2}b+a{b}^2$.

• A. True
• B. False

Answer 1

The correct option is B False

Given $a,b\in +R⟹a\ge 0,b\ge 0$ and $a\ne b$

Therefore, $a+b>0$

$a^3+b^3-a^{2}b-a{b}^2=a^3-a^{2}b-a{b}^2+b^3$

$=a^2(a-b)-b^2(a-b)$

$=(a^2-b^2)(a-b)$

$=(a+b)(a-b)(a-b)$

$=(a+b)(a-b{)}^2>0$

Because $(a-b{)}^2$ is always $\ge 0$ (since, square term is always positive) and

$a+b>0$

therefore, $a^3+b^3-a^{2}b-a{b}^2>0$

$⟹a^3+b^3>a^{2}b+a{b}^2$