Menu Close

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




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

Leave a Reply

Your email address will not be published. Required fields are marked *