Question #68e59

2 Answers
xurongna_
Sep 12, 2017

#(2a^3 + b^3) / (a^3+2b^3)#

Explanation:

#((a+b)^3 + a^3) / ((a+b)^3 + b^3)#

A rule of exponents is that variables with an operation within the bracket are raised to the exponent. E.g.
#(p+q)^2#
#=p^2 + q^2#

So here, you apply the same rule:
#=(a^3 + b^3 + a^3) / (a^3 + b^3 + b^3)#
#=(2a^3 + b^3) / (a^3 + 2b^3)#

You can't factorise the expression because of the +

Hope this helps!

Jim G.
Sep 12, 2017

#(2a+b)/(2b+a)#

Explanation:

#(a+b)^3+a^3" is a "color(blue)"sum of cubes"#

#•color(white)(x)x^3+y^3=(x+y)(x^2-xy+y^2)larrcolor(blue)" factors"#

#"here "x=a+b" and "y=a#

#rArr(a+b)^3+a^3#

#=(a+b+a)((a+b)^2-a(a+b)+a^2)#

#=(2a+b)(a^2+2ab+b^2cancel(-a^2)-abcancel(+a^2))#

#=(2a+b)(a^2+b^2+ab)#

#color(blue)"Similarly for denominator"#

#"here "x=a+b" and " y=b#

#rArr(a+b)^3+b^3#

#=(a+b+b)((a+b)^2-b(a+b)+b^2)#

#=(2b+a)(a^2+2ab+b^2-abcancel(-b^2)cancel(+b^2))#

#=(2b+a)(a^2+b^2+ab)#

#"Putting it together we obtain"#

#((2a+b)cancel((a^2+b^2+ab)))/((2b+a)cancel((a^2+b^2+ab)))#

#=(2a+b)/(2b+a)#

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