Question

3a4-48b4 factorize this equation

Answer 1

Given :

Factorize

3a⁴ - 48b⁴

Solution :

Solve this question by applying identity

Solve this question by applying identitya² - b² = (a+b)(a-b)

= 3a⁴ - 48b⁴

Take 3 as a common

= 3 (a⁴ - 16b⁴)

= 3 [ (a²)² - (4b²)²]

= 3 [ (a²+4b²)(a²-4b²) ]

= 3 (a²+4b²)[ (a)²-(2b)² ]

= 3 (a²+4b²)[ (a+2b)(a-2b) ]

= 3 (a²+4b²)(a+2b)(a-2b)

Important identities :

★ (a+b)² = a² + b² + 2ab

★ (a-b)² = a² + b² - 2ab

★ (a-b)³ = a³ - b³ - 3ab(a-b)

★ (a+b)³ = a³ + b³ + 3ab(a+b)

★ a² - b² = (a+b)(a-b)

★ a³ + b³ = (a+b)(a²- ab + b²)

★ a³ - b³ = (a-b)(a² + ab + b²)

Answer 2

$\huge\underline\mathrm{GivEn:-}$

3a⁴ - 48b⁴

$\huge\underline\mathrm{To Find:-}$

↪ Factorize

$\huge\underline\mathrm{SoLuTion:-}$

Using Identity, a² - b² = (a + b)(a - b), solve the above equation

$\implies$ 3a⁴ - 48b⁴

$\implies$ 3(a⁴ - 16b⁴)

$\implies$ 3[(a²)² - (4b²)²]

$\implies$ 3[(a² + 4b²)(a² - 4b²)]

$\implies$ 3[(a² + 4b²)(a + 2b)(a - 2b)]

$\huge\underline\mathrm{ThAnKs...}$