MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Expansion of (x^2 - y^2)

Factorise : ...

Question

Factorise : $(a^{2} - b^{2}) (a^{2} + a b + b^{2}) - a^{3} + b^{3}$

Open in App

Solution

We have,
$(a^{2} - b^{2}) (a^{2} + a b + b^{2}) - a^{3} + b^{3}$
$\Rightarrow (a^{2} - b^{2}) (a^{2} + a b + b^{2}) - (a^{3} - b^{3})$
$\Rightarrow (a - b) (a + b) (a^{2} + a b + b^{2}) - (a - b) (a^{2} + a b + b^{2})$
$\Rightarrow (a + b - 1) (a - b) (a^{2} + a b + b^{2})$

Hence, this is the answer.

Suggest Corrections

thumbs-up

1

Similar questions

csc θ - sin θ = a^{3}

sec θ - cos θ = b^{3}

a^{2} b^{2} (a^{2} + b^{2}) = 1

csc A - sin A = a^{3}

sec A - cos A = b^{3}

a^{2} b^{2} (a^{2} + b^{2}) = 1

csc θ - sin θ = a^{3}

sec θ - cos θ = b^{3}

a^{2} b^{2} (a^{2} + b^{2}) = 1

a - a^{3} + b^{3} - b

cosec x - \sin x = a^{3}, \sec x - \cos x = b^{3}

, then prove that

a^{2} b^{2} (a^{2} + b^{2}) = 1

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

x^2 - y^2

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Expansion of (x^2 - y^2)

Standard VIII Mathematics

Join BYJU'S Learning Program

CrossIcon