Question

show that:

${(2p - q)}^{2} + 8pq = {(2p + q)}^{2}$
using these identities please send me the answer :-
${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2(a)(b)$
${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2 (a)(b)$
${a}^{2} - {b}^{2} = ( a+ b)(a-b )$
$(x + a)(x + b) = x(x + b) + a(x + b)$
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Answer 1

Answer:

See the explanation

Step-by-step explanation:

$\text{LHS}=(2p-q)^2+8pq$

       $=(2p)^2+q^2-2(2p)(q)+8pq$

       $=4p^2+q^2-4pq+8pq\\=4p^2+q^2+4pq$

$\text{RHS}=(2p+q)^2$

       $=(2p)^2+q^2+2(2p)(q)$

       $=4p^2+q^2+4pq$

Clearly $\text{LHS}=\text{RHS}$, hence proved