Question

x^6 -1 factorise it...? ​

Answer 1

hiii

x^6-1=(x^3)2-(1)^2

and we know a^2-b^2=(a+b)(a-b)

so

here

(x^3-1)(x^3+1)

also,a^3-b^3=(a-b)(a^2+b^2+ab)

and a^3+b^3=(a+b)(a^2+b^2-ab)

so,

using that identity

we can further expand it as

(x-1)(x^2+1+x)(x+1)(x^2+1-x)

easiest question✔️✔️

Answer 2

To factorise:

• x^6 - 1 = 0

Solution:

[Using algebraic identity

• a^2 - b^2 = (a + b) (a - b)]

➡ x^6 - 1

➡ ((x)^3)^2 - (1)^2

➡ (x^3 + 1) (x^3 - 1)

[Now, Using algebraic identities

• a^3 + b^3 = (a + b) (a^2 + b^2 - ab)

and

• a^3 - b^3 = (a - b) (a^2 + b^2 + ab)]

➡ (x^3 + 1^3) (x^3 - 1^3)

➡ (x + 1) (x^2 + 1^2 - x(1)) (x - 1) (x^2 + 1^2 + x(1))

➡ (x + 1) (x^2 - x + 1) (x - 1) (x^2 + x + 1)

Factorised.