How do you simplify #(64x^3+1)/( 4x^2-100) * (4x+20)/(64x^2-16x+4)#?

1 Answer
Jul 18, 2015

Answer:

#(64x^3+1)/(4x^2-100)*(4x+20)/(64x^2-16x+4)#

#=(4x+1)/(4(x-5))#

with exclusion #x != -5#

Explanation:

Use sum of cubes identity: #a^3+b^3 = (a+b)(a^2-ab+b^2)#

Use difference of squares identity: #a^2-b^2 = (a-b)(a+b)#

#(64x^3+1)/(4x^2-100)*(4x+20)/(64x^2-16x+4)#

#=(((4x)^3+1^3)*4(x+5))/(4(x^2-5^2)*4((4x)^2-4x+1))#

#=(4(4x+1)((4x)^2-4x+1)(x+5))/(16(x-5)(x+5)((4x)^2-4x+1))#

#=(4x+1)/(4(x-5))#

with exclusion #x != -5#