Question

$15(a^3{b}^2{c}^2-a^2{b}^3{c}^2+a^2{b}^2{c}^3)÷3abc$

• A. $abc(a-b+c)$
• B. $5abc(a+b+c)$
• C. $5abc(a-b+c)$
• D. None

Answer 1

The correct option is C $5abc(a-b+c)$

We divide the given polynomial $15(a^3{b}^2{c}^2-a^2{b}^3{c}^2+a^2{b}^2{c}^3)$ by the monomial $3abc$ as shown below:

$\frac{15(a^3{b}^2{c}^2-a^2{b}^3{c}^2+a^2{b}^2{c}^3)}{3abc}=\frac{15a^3{b}^2{c}^2-15a^2{b}^3{c}^2+15a^2{b}^2{c}^3}{3abc}=\frac{15a^3{b}^2{c}^2}{3abc}-\frac{15a^2{b}^3{c}^2}{3abc}+\frac{15a^2{b}^2{c}^3}{3abc}=\left(\frac{3\times 5\times a\times a\times a\times b\times b\times c\times c}{3abc}\right)-\left(\frac{3\times 5\times a\times a\times b\times b\times b\times c\times c}{3abc}\right)+\left(\frac{3\times 5\times a\times a\times b\times b\times c\times c\times c}{3abc}\right)$

$=5a^{2}bc-5a{b}^{2}c+5ab{c}^2=5abc(a-b+c)$

Hence, $\frac{15(a^3{b}^2{c}^2-a^2{b}^3{c}^2+a^2{b}^2{c}^3)}{3abc}=5abc(a-b+c)$.