Factorise p⁴ + q⁴ + p²q².

Solution:

Given, p⁴ + q⁴ + p²q²

Here, p⁴ + q⁴ + p²q² can be written as (p²)² + (q²)² + 2p²q² - p²q²

∴ p⁴ + q⁴ + p²q²

= (p²)² + (q²)² + 2p²q² - p²q²

= (p² + q²)² - p²q²

= (p² + q²)² - (pq)²

Using the identity: a² - b² = (a + b) (a - b)

We get, (p² + q² + pq) (p² + q² - pq)

✦ Try This: Factorise: x⁸ + y⁸ + x⁴y⁴

Given, x⁸ + y⁸ + x⁴y⁴

Here, x⁸+ y⁸ + x⁴y⁴ can be written as (x⁴)² + (y⁴)² + 2x⁴y⁴ - x⁴y⁴

∴ x⁸ + y⁸ + x⁴y⁴

= (x⁴)² + (y⁴)² + 2x⁴y⁴ - x⁴y⁴

= (x⁴ + y⁴)² - (x²y²)²

Using the identity: a² - b² = (a + b) (a - b)

= (x⁴ + y⁴ + x²y²)(x⁴ + y⁴ - x²y²)

☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9

NCERT Exemplar Class 8 Maths Chapter 7 Problem 119

Factorise p⁴ + q⁴ + p²q².

Summary:

Factorising p⁴ + q⁴ + p²q² we get, (p² + q² + pq) (p² + q² - pq)

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