Question

${(3pq+3qr)}^2-18p{q}^{2}r$ = .

• A. $9p{q}^2+9q{r}^2-18p$
• B. $9p^{2}q+9p{r}^2-9pq$
• C. $9p^2{q}^2+9q^2{r}^2$
• D. $9p^2{q}^2+9q{r}^2+9p$

Answer 1

The correct option is C $9p^2{q}^2+9q^2{r}^2$

Given: ${(3pq+3qr)}^2-18p{q}^{2}r$

Consider ${(3pq+3qr)}^2$
Expand this using the identity
$({(a+b)}^2=a^2+2ab+b^2)$
${(3pq+3qr)}^2=9{\left(p\right)}^2{\left(q\right)}^2+9{\left(q\right)}^2{\left(r\right)}^2+18p{q}^{2}r$

${(3pq+3qr)}^2-18p{q}^{2}r=9p^2{q}^2+9q^2{r}^2+18p{q}^{2}r-18p{q}^{2}r$
$=9p^2{q}^2+9q^2{r}^2$