Question

find the following product using suitable identity:(¾p²+⅔q²)(¾p²+⅔q²)
$plz \: dont \: give \: silly \: answers \: or \: i \: will \: report \: the \: answer$
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Answer 1

Given :

$\bf\longmapsto{\bigg( \dfrac{3}{4}p^{2} + \dfrac{2}{3}q^{2}\bigg) \bigg( \dfrac{3}{4}p^{2} + \dfrac{2}{3}q^{2} \bigg)}$

Identity :

$\boxed{\sf{(a + b)^{2} = a^{2} + b^{2} + 2ab)}}$

Now, As we know

Here,

a = $\tt\longmapsto{\bigg( \dfrac{3}{4}p^{2}\bigg)}$

b = $\tt\longmapsto{\bigg( \dfrac{2}{3}q^{2}\bigg)}$

Solution :

Substituting the respective values in it, we get :

$\tt\longmapsto{\bigg( \dfrac{3}{4}p^{2} + \dfrac{2}{3}q^{2}\bigg) \bigg( \dfrac{3}{4}p^{2} + \dfrac{2}{3}q^{2} \bigg)}$

$\tt\longmapsto{\bigg( \dfrac{3}{4}p^{2} + \dfrac{2}{3}q^{2} \bigg)^{2}}$

$\tt\longmapsto{\bigg( \dfrac{3}{4}p^{2}\bigg)^{2} + \bigg( \dfrac{2}{3}q^{2}\bigg)^{2} + 2 \times \bigg( \dfrac{3}{4}p^{2}\bigg) \times \bigg( \dfrac{2}{3}q^{2}\bigg)}$

$\tt\longmapsto{\bigg( \dfrac{9}{16}p^{4}\bigg) + \bigg( \dfrac{4}{9}q^{4}\bigg) + 2 \times \bigg( \dfrac{6}{12}p^{2}q^{2}\bigg)}$

$\tt\longmapsto{\bigg( \dfrac{9}{16}p^{4}\bigg) + \bigg( \dfrac{4}{9}q^{4}\bigg) + \bigg( \dfrac{12}{12}p^{2}q^{2}\bigg)}$

$\tt\longmapsto{\bigg( \dfrac{9}{16}p^{4}\bigg) + \bigg( \dfrac{4}{9}q^{4}\bigg) + \bigg( \dfrac{\cancel 12}{\cancel 12}p^{2}q^{2}\bigg)}$

$\tt\longmapsto{\bigg( \dfrac{9}{16}p^{4}\bigg) + \bigg( \dfrac{4}{9}q^{4}\bigg) + \bigg(1p^{2}q^{2}\bigg)}$

Therefore ,

$\boxed{\bf{\bigg( \dfrac{9}{16}p^{4}\bigg) + \bigg( \dfrac{4}{9}q^{4}\bigg) + \bigg(p^{2}q^{2}\bigg)}}$

Some Important Identities :

★ (a + b)² = a² + 2 ab + b²

 

★(a - b)² = (a + b)² - 4 ab

★ a² - b² = (a + b) (a - b)

 

★ (x + a) (x + b) =x² + (a + b) x + ab

★ (a + b)² = (a - b)² + 4 ab

★ (a - b)² = a² - 2 ab + b²

★ (a - b)³= a³- 3a²b + 3ab² - b³

★ (a - b)³ = a³- b³-3ab (a-b)

★ a³+ b³ = (a + b) (a²-ab + b²)

★ a³- b³= (a - b)(a²+ ab + b²)

★ a³- b³= (a - b)³ + 3ab(a-b)

★ (a + b + c)²= a²+ b²+ c² + 2ab + 2bc + 2ca

★ (a + b - c)² = a²+ b²+ c² + 2ab - 2bc -2ca

★ (a - b + c)²= a² + b²+c²-2ab -2bc +2ca

★ (a - b - c)²= a²+b²+c²-2ab +2bc -2ca

★a² + b² = (a - b)² + 2ab

★  (a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab²+3b²c + 3ac² + 3bc² + 6abc

★ (a - b - c)³ = a³ - b³ - c³ - 3a²b - 3a²c + 3ab² + 3b²c + 3ac² - 3bc² + 6abc

Answer 2

Answer:

(9/16p⁴)+(4/9q⁴)+(p²q²)....hope it is helpful to you... please mark me as BRAINLIEST...