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  • Factorise: (i) 1+64x^{3} (ii) a^{3}-2\sqrt{2}b^{3}

Factorise:

(i) 1+64x^{3}

(ii) a^{3}-2\sqrt{2}b^{3}

 

Answers (1)

(i) (1+4x)(1+16x^{2}-4x)

Given 1+64^{3}

(1)^{3}+(4x)^{3}

we know that  a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)

so (1)^{3}+\left (4x \right )^{3}=(1+4x)((1)^{2}+\left (4x \right )^{2}-(1)\left (4x \right ))

(1+4x)(1+16x^{2}-4x)

(ii) (a-\sqrt{2}b)(a^{2}+2b^{2}+\sqrt{2}ab)

Given a^{3}-2\sqrt{2}b^{3}

This can be written as

(a)^{3}-(\sqrt{2}b)^{3}

we know that  a^{3}-b^{3}=(a+b)(a^{2}+b^{2}+ab)

so (a)^{3}-(\sqrt{2}b)^{3}=(a-\sqrt{2}b)(a^{2}+(\sqrt{2}b)^{2}+\sqrt{2}ab)

(a-\sqrt{2}b)(a^{2}+2b^{2}+\sqrt{2}ab)

Posted by

infoexpert24

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