Question Number 27449 by rohith siddharth last updated on 07/Jan/18
$${factorise}\:{a}^{\mathrm{4}} −\left({b}+{c}\right)^{\mathrm{4}} \\ $$
Answered by $@ty@m last updated on 07/Jan/18
$${Let}\:{b}+{c}={d} \\ $$$${a}^{\mathrm{4}} −{d}^{\mathrm{4}} \\ $$$$=\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({d}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$=\left({a}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} \right) \\ $$$$=\left({a}+{d}\right)\left({a}−{d}\right)\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} \right) \\ $$$$=\left({a}+{b}+{c}\right)\left({a}−{b}−{c}\right)\left\{{a}^{\mathrm{2}} +\left({b}+{c}\right)^{\mathrm{2}} \right\} \\ $$
Commented by rohith siddharth last updated on 07/Jan/18
$$\mathrm{I}\:\mathrm{want}\:\mathrm{answer}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step} \\ $$
Commented by $@ty@m last updated on 07/Jan/18
$${Is}\:{it}\:{okay}\:{now}? \\ $$
Commented by rohith siddharth last updated on 07/Jan/18
$$\mathrm{ok}\:\mathrm{thank}\:\mathrm{you} \\ $$
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 ) Recall also p^2 + q^2 = (p + q)^2 − 2pq = (a + b + c)(a − b − c)(a^2 + (b + c)^2 ) = (a + b + c)(a − b − c)[(a + b + c)^2 − 2a(b + c)]](https://www.tinkutara.com/question/Q27453.png)
$$\mathrm{Recall}\:\mathrm{that}\:{p}^{\mathrm{2}} \:−\:{q}^{\mathrm{2}} \:=\:\left({p}\:+\:{q}\right)\left({p}\:−\:{q}\right) \\ $$$$ \\ $$$${a}^{\mathrm{4}} \:−\:\left({b}\:+\:{c}\right)^{\mathrm{4}} \:=\:\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\:\left(\left({b}\:+\:{c}\right)^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}^{\mathrm{2}} \:−\:\left({b}\:+\:{c}\right)^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+\:\left({b}\:+\:{c}\right)^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}\:+\:{b}\:+\:{c}\right)\left({a}\:−\:\left({b}\:+\:{c}\right)\right)\left({a}^{\mathrm{2}} \:+\:\left({b}\:+\:{c}\right)^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}\:+\:{b}\:+\:{c}\right)\left({a}\:−\:{b}\:−\:{c}\right)\left({a}^{\mathrm{2}} \:+\:\left({b}\:+\:{c}\right)^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{R}{e}\mathrm{call}\:\mathrm{also}\:{p}^{\mathrm{2}} \:+\:{q}^{\mathrm{2}} \:=\:\:\left({p}\:+\:{q}\right)^{\mathrm{2}} \:−\:\mathrm{2}{pq} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}\:+\:{b}\:+\:{c}\right)\left({a}\:−\:{b}\:−\:{c}\right)\left({a}^{\mathrm{2}} \:+\:\left({b}\:+\:{c}\right)^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}\:+\:{b}\:+\:{c}\right)\left({a}\:−\:{b}\:−\:{c}\right)\left[\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{2}} \:−\:\mathrm{2}{a}\left({b}\:+\:{c}\right)\right] \\ $$
Commented by rohith siddharth last updated on 07/Jan/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{bro} \\ $$