Question

Show that (8pq+5q)²-(8pq-5q)²=160pq²​

Answer 1

Step-by-step explanation:

Replace m by q here...

Let x be any positive integer

Then x = 5q or x = 5q+1 or x = 5q+4 for integer x.

If x = 5q, x2 = (5q)2 = 25q2 = 5(5q2) = 5n (where n = 5q2 )

If x = 5q+1, x2 = (5q+1)2 = 25q2+10q+1 = 5(5q2+2q)+1 = 5n+1 (where n = 5q2+2q )

If x = 5q+4, x2 = (5q+4)2 = 25q2+40q+16 = 5(5q2 + 8q + 3)+ 1 = 5n+1 (where n = 5q2+8q+3 )

∴in each of three cases x2 is either of the form 5q or 5q+1 or 5q+4 and for integer q.

Answer 2

• To Prove $\rm{\left(8pq+5q\right)^2-\left(8pq-5q\right)^2=160pq^2}$ , First We Expand LHS that is $\rm{\left(8pq+5q\right)^2-\left(8pq-5q\right)^2}$ and if it equals RHS that is $\rm{160pq^2}$, Then we are able to Prove $\rm{\left(8pq+5q\right)^2-\left(8pq-5q\right)^2=160pq^2}$

Taking LHS (Left Hand Side)

$\rm{\left(8pq+5q\right)^2-\left(8pq-5q\right)^2}$

$\rm{=64p^2q^2+80pq^2+25q^2-\left(8pq-5q\right)^2}$

$\rm{=64p^2q^2+80pq^2+25q^2-\left(64p^2q^2-80pq^2+25q^2\right)}$

$\rm{=64p^2q^2+80pq^2+25q^2-64p^2q^2+80pq^2-25q^2}$

$\rm{=64p^2q^2-64p^2q^2+80pq^2+80pq^2+25q^2-25q^2}$

$\rm{=80pq^2+80pq^2+25q^2-25q^2}$

$\rm{=160pq^2+25q^2-25q^2}$

$\rm{=160pq^2}$

Which is Equal to RHS (Right Hand Side)

And Therefore We Showed that $\rm{\left(8pq+5q\right)^2-\left(8pq-5q\right)^2=160pq^2}$