Question

(7pq + 4ab)2...............​

Answer 1

Correct Question:-

$\sf(7pq + 4ab) {}^{2}$

Using identity 1 :- $\boxed{\sf{(a+b)^2=a^2+b^2+2ab}}$

$\sf(7pq + 4ab)\\ = \sf{(7pq)}^{2} + {(4ab)}^{2} + 2 \times 7pq \times 4ab \\ = \sf49 {p}^{2} {q}^{2} + 16 {a}^{2} {b}^{2} + 56abpq$

Hope you've understood the sum!

Answer 2

$\sf{(7pq + 4 {ab})^{2} }$

Using the formula:

$\tt{ {a}^{2} + {b}^{2} = {a}^{2} + 2ab + {b}^{2} }$

Here, a = 7pq and b = 4ab

$\tt{7pq + 4ab}$

$\tt{ = ({7pq})^{2} + 2.7pq.4ab + {(4ab}^{2} ) }$

$\tt{ = 46{p}^{2} {q}^{2} + 56abpq + 16 {a}^{2} {b}^{2} }$

$\huge\underline\mathfrak\red{thankyou}$