Question

desenvolva[(2p+3)² + (2p-3)²]².

Answer 1

[(2p+3)² + (2p-3)²]²=

[4p²+12p+9+4p²-12p+9]²

(8p² +18)²=

$64p^4+288p^2+324$

Answer 2

$[(2p+3)^2+(2p-3)^2]^2$

Primeiramente, veja que:

$\rhd$ $(2p+3)^2=(2p)^2+2\cdot(2p)\cdot3+3^2=4p^2+12p+9$

$\rhd$ $(2p-3)^2=(2p)^2-2\cdot(2p)\cdot3+3^2=4p^2-12p+9$

Assim:

$[(2p+3)^2+(2p-3)^2]^2=[(4p^2+12p+9)+(4p^2-12p+9)]^2=(8p^2+18)^2$

$\bullet$ $(8p^2+18)^2=(8p^2)^2+2\cdot(8p^2)\cdot18+18^2=64p^4+288p^2+324$

Logo:

$\boxed{[(2p+3)^2+(2p-3)^2]^2=64p^4+288p^2+324}$