Question

show that (4p^2-3q)^2+48p^2q =(4p^2+3q^2)​

Answer 1

Given:

The statement is

$(4p^2-3q)^2+48p^2q=(4p^2+3q^2)$

To prove:

The given statement.

Solution:

We have,

$(4p^2-3q)^2+48p^2q=(4p^2+3q^2)$

Now,

$LHS=(4p^2-3q)^2+48p^2q$

Using $(a-b)^2=a^2+b^2-2ab$, we get

$LHS=(4p^2)^2+(3q)^2-2(4p^2)(3q)+48p^2q$

$LHS=(4p^2)^2+(3q)^2-24p^2q+48p^2q$

$LHS=(4p^2)^2+(3q)^2+24p^2q$

It can be rewritten as

$LHS=(4p^2)^2+(3q)^2+2(4p^2)(3q)$

Using $(a+b)^2=a^2+b^2+2ab$, we get

$LHS=(4p^2+3q^2)$

$LHS=RHS$

Hence proved.