R. D. Sharma Solutions for Chapter: Algebraic Identities, Exercise 7: MULTIPLE CHOICE QUESTIONS (MCQs)

If $a+b+c=0$, then $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=$

If $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0$, then

If $a+b+c=9$ and $ab+bc+ca=23$, then $a^3+b^3+c^3-3abc=$

Simplify $\frac{{\left(a^2-b^2\right)}^3+{\left(b^2-c^2\right)}^3+{\left(c^2-a^2\right)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}$

The product $(a+b)(a-b)\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)$ is equal to

The product $\left(x^2-1\right)\left(x^4+x^2+1\right)$ is equal to

If $\frac{a}{b}+\frac{b}{a}=1$, then $a^3+b^3=$

If $49a^2-b=\left(7a+\frac{1}{2}\right)\left(7a-\frac{1}{2}\right)$, then the value of $b$ is