Solve (sic) #(x^4-a^4)/(x-a)# ?

Question

417 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T02:29:32

Expression #=(x^2+a^2)(x+a) # #[x!=a]#

NB: By "solve" I have assumed meant "simplify".

Expression #= (x^4-a^4)/(x-a)#

Remember that: #p^2-q^2 = (p+q)(p-q)#

Hence, Expression #= ((x^2+a^2)(x^2-a^2))/(x-a)#

#= ((x^2+a^2)(x+a)(x-a))/(x-a)#

So long as #x!=a# we can cancel #(x-a)#

#= ((x^2+a^2)(x+a)cancel((x-a)))/cancel((x-a))# #[x!=a]#

#= (x^2+a^2)(x+a) #