Question

(2a – 5b) (2a + 5b) ( 4a2 + 25 b2)​

Answer 1

GIVEN :-

• (2a - 5b) (2a + 5b) (4a² + 25b²)

TO FIND :-

• The value of (2a - 5b) (2a + 5b) (4a² + 25b²).

SOLUTION :-

Firstly , Let's divide the equation into two parts,

⇒ [(2a - 5b) (2a + 5b)] [(4a² + 25b²)]

Now by using identity (a + b)(a - b) = a² - b².

⇒ [ (2a)² - (5b)² ] [ 4a² + 25b² ]

⇒ (4a² - 25b²)(4a² + 25b²)

Now by using identity (a + b)(a - b) = a² - b².

⇒ (4a²)² - (25b²)²

⇒ 16a⁴ - 625b⁴

⇒ (2a - 5b) (2a + 5b) (4a² + 25b²) = 16a⁴ - 625b⁴.

Hence the value of (2a - 5b) (2a + 5b) (4a² + 25b²) is 16a⁴ - 625b⁴.

Answer 2

$\underline{\purple{\frak{ \: \: Given:- \: \: }}}$

• $\sf (2a-5b)(2a+5b)(4a^2 + 25b^2)$

$\underline{\purple{\frak{ \: \: Find:- \: \: }}}$

• $\sf Value\:of\:(2a-5b)(2a+5b)(4a^2 + 25b^2)$

$\underline{\purple{\frak{ \: \: Solution:- \: \: }}}$

we, have

$\dashrightarrow\sf (2a-5b)(2a+5b)(4a^2 + 25b^2)$

$\dashrightarrow\sf (2a+5b)(2a - 5b)(4a^2 + 25b^2)$

$\dashrightarrow\sf \{(2a)^2 - (5b)^2 \}(4a^2 + 25b^2) \quad \bigg\lgroup\because (a+b)(a-b) = a^2 - b^2\bigg\rgroup$

$\dashrightarrow\sf (4a^2 - 25b^2)(4a^2 + 25b^2)$

$\dashrightarrow\sf (4a^2 + 25b^2) (4a^2 - 25b^2)$

$\dashrightarrow\sf (4a^2)^2 - (25b^2)^2\quad \bigg\lgroup\because (a+b)(a-b) = a^2 - b^2\bigg\rgroup$

$\dashrightarrow\sf 16a^4 - 625b^4$

$\underline{\boxed{\sf\therefore (2a-5b)(2a+5b)(4a^2+25b^2) = 16a^4 - 625b^4}}$