Question

factorise 1 a2+b2+2bc-2ca-2ab 2 nd x8-y8

Answer 1

We can write

a² + b² + 2bc - 2ca - 2ab, as

a² + b² + c² + 2bc - 2ca - 2ab - c²

Now using the identity

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

Here, 2ca and 2ab are negative and both has a common.

=> a is negative.

Hence,

a² + b² + c² + 2bc - 2ca - 2ab - c²

= (b - a + c)² - (c)²

Now using the identity

a² - b² = (a + b)(a - b)

=> (b - a + c)² - (c)²

= (b - a + c + c)(b - a + c - c)

= (b - a + 2c)(b - a)




2) x^8 - y^8

=> (x⁴)² - (y⁴)²

=> (x⁴ + y⁴)(x⁴ - y⁴)

=> (x⁴ + y⁴)[ (x²)² - (y²)²]

=> (x⁴ + y⁴)(x² + y²)(x² - y²)


=> (x⁴ + y⁴)(x² + y²)(x + y)(x - y)


Hope it helps dear friend ☺️