Question

Addition of $\frac{a^2}{2}+\frac{b^3}{3}-\frac{c^3}{4}$,
$\frac{{2a}^2}{3}+\frac{{3b}^3}{4}-\frac{{4c}^3}{5}$ and $a^2+b^3+c^3$ is ______.

• A. $\frac{13}{6}a+\frac{25}{12}{b}^3-\frac{1}{20}{c}^3$
• B. $\frac{13}{6}{a}^2-\frac{1}{20}{b}^3+\frac{25}{12}{c}^3$
• C. $\frac{13}{6}{a}^2+\frac{25}{12}{b}^3-\frac{1}{20}{c}^3$
• D. $\frac{13}{6}{a}^2-\frac{25}{12}{b}^3+\frac{1}{20}{c}^3$

Answer 1

The correct option is C $\frac{13}{6}{a}^2+\frac{25}{12}{b}^3-\frac{1}{20}{c}^3$

According to question-

$=(\frac{a^2}{2}+\frac{b^3}{3}-\frac{c^3}{4})+(\frac{2a^2}{3}+\frac{3b^3}{4}-\frac{4c^3}{5})+(a^2+b^3+c^3)$

Adding same term,

$=(\frac{a^2}{2}+\frac{2a^2}{3}+a^2)+(\frac{b^3}{3}+\frac{3b^3}{4}+b^2)+(\frac{-c^3}{4}-\frac{4c^3}{5}+c^3)$

$=\frac{3a^2+4a^2+6a^2}{6}+\frac{4b^3+9b^3+12b^3}{12}+\frac{-5c^3-16c^3+20c^3}{20}$

$=\frac{13}{6}{a}^2+\frac{25}{12}{b}^3-\frac{1}{20}{c}^3$