factorise-a-4-b-c-4-

Question Number 27449 by rohith siddharth last updated on 07/Jan/18

f a c t o r i s e a^{4} - {(b + c)}^{4}

Answered by $@ty@m last updated on 07/Jan/18

L e t b + c = d

a^{4} - d^{4}

= {(a^{2})}^{2} - {(d^{2})}^{2}

= (a^{2} - d^{2}) (a^{2} + d^{2})

= (a + d) (a - d) (a^{2} + d^{2})

= (a + b + c) (a - b - c) {a^{2} + {(b + c)}^{2}}

Commented by rohith siddharth last updated on 07/Jan/18

I want answer step by step

Commented by $@ty@m last updated on 07/Jan/18

I s i t o k a y n o w ?

Commented by rohith siddharth last updated on 07/Jan/18

ok thank you

Answered by Joel578 last updated on 07/Jan/18

Recall that p^{2} - q^{2} = (p + q) (p - q)

a^{4} - {(b + c)}^{4} = {(a^{2})}^{2} - {({(b + c)}^{2})}^{2}

= (a^{2} - {(b + c)}^{2}) (a^{2} + {(b + c)}^{2})

= (a + b + c) (a - (b + c)) (a^{2} + {(b + c)}^{2})

= (a + b + c) (a - b - c) (a^{2} + {(b + c)}^{2})

R e call also p^{2} + q^{2} = {(p + q)}^{2} - 2 p q

= (a + b + c) (a - b - c) (a^{2} + {(b + c)}^{2})

= (a + b + c) (a - b - c) [{(a + b + c)}^{2} - 2 a (b + c)]

Commented by rohith siddharth last updated on 07/Jan/18

thank you bro