Which of the following is equal to $X$ ?
(a) $x^{\frac{12}{7}}-x^{\frac{5}{7}}$
(b) $\sqrt[12]{\left(x^{4}\right)^{\frac{1}{3}}}$
(c) $\left(\sqrt{x^{3}}\right)^{\frac{2}{3}}$
(d) $x^{\frac{12}{7}} \times x^{\frac{7}{12}}$
(c)
(a) $x^{\frac{12}{7}}-x^{\frac{5}{7}}=x^{\frac{5}{7}+1}-x^{\frac{5}{7}}$
$=x^{\frac{5}{7}} \cdot x-x^{\frac{5}{7}}$ $\left[\because a^{m+n}=a^{m} a^{n}\right]$
(b) $\sqrt[12]{\left(x^{4}\right)^{\frac{1}{3}}}=\left(\left(x^{4}\right)^{\frac{1}{3}}\right)^{\frac{1}{12}}$ $\left.[\because m]{a}=a^{1 / m}\right]$
$=x^{4 \times \frac{1}{3} \times \frac{1}{12}}=x^{\frac{1}{9}}$ $\left[\because\left\{\left(x^{m}\right)^{n}\right\}^{p}=x^{m n p}\right]$
$\neq x$
(c) $\left(\sqrt{x^{3}}\right)^{\frac{2}{3}}=\left(x^{\frac{3}{2}}\right)^{\frac{2}{3}}=x^{\frac{3}{2} \times \frac{2}{3}}$ $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$
$=x$
(d) $x^{\frac{12}{7}} \times x^{\frac{7}{12}}=x^{\frac{12}{7}+\frac{7}{12}}$ $\left[\because a^{m} \cdot a^{n}=a^{m+n}\right]$
$=x^{\frac{144+49}{84}}=x^{\frac{193}{84}} \neq x$