How do you simplify #[(x-a)+(2a^2)/(x+a)] * [((1/x^2)-(1/a^2))] div[((x^3)-((ax^3+a^4)/(x+a)))]#?

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Answer 1 · 2017-03-30T04:33:38

#-1/(x^2a^2).#

The Exp.#=[{(x-a)(x+a)+2a^2}/(x+a)]*[(a^2-x^2)/(x^2a^2)]-:[x^3-{(a(x^3+a^3))/(x+a)}],#

#=[{(x^2-a^2)+2a^2}/(x+a)]*[{(a+x)(a-x)}/(x^2a^2)]-:[x^3-{a(x+a)(x^2-xa+a^2)}/(x+a)],#

#=(x^2+a^2)*{(a-x)/(x^2a^2)}-:[x^3-a(x^2-xa+a^2)],#

#={(x^2+a^2)(a-x)}/(x^2a^2)-:{x^3-ax^2+xa^2-a^3},#

#={(x^2+a^2)(a-x)}/(x^2a^2)-:{(x^3-a^3)-xa(x-a)},#

#={(x^2+a^2)(a-x)}/(x^2a^2)-:{(x-a)(x^2+xa+a^2)-xa(x-a)},#

#={(x^2+a^2)(a-x)}/(x^2a^2)-:{(x-a)(x^2+xa+a^2-xa)},#

#={(x^2+a^2)(a-x)}/(x^2a^2)-:{(x-a)(x^2+a^2)},#

#={(x^2+a^2)(a-x)}/(x^2a^2)xx1/{-(a-x)(x^2+a^2)},#

# rArr" The Exp.="-1/(x^2a^2).#

Enjoy Maths.!

Answer 2 · 2017-03-30T08:52:15

#-1/(a^2x^2)#

Calling

#f_1 = x-a+(2a^2)/(x+a) = (x^2-a^2+2a^2)/(x+a)=(x^2+a^2)/(x+a)#

#f_2=1/x^2-1/a^2=(a^2-x^2)/(a^2 x^2)#

#f_3 = x^3-(ax^3+a^4)/(x+a)=(x^4+ax^3-ax^3-a^4)/(x+a)=(x^4-a^4)/(x+a)#

we have

#(f_1 f_2)/(f_3)=((x^2+a^2)/(x+a))((a^2-x^2)/(a^2 x^2))((x+a)/(x^4-a^4))=-1/(a^2x^2)#