#(a^2-b^2)/(a-b)-(a^3-b^3)/(a^2-b^2)=?#

i need how to start this question ty!

2 Answers
Jim G.
May 11, 2018

#(ab)/(a+b)#

Explanation:

#"simplify numerators/denominators using"#

#color(blue)"difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"and "color(blue)"difference of cubes"#

#•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)#

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

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

#"multiply and divide "(a+b)" by "(a+b)#

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

#"the fractions now have a "color(blue)"common denominator"#
#"so add/subtract numerators leaving the denominator"#

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

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

Mahek ☮
May 11, 2018

#color(magenta)(=>(ab)/(a+b)#

Explanation:

#∙# #color(blue)[(a^2−b^2)]/(a−b)−color(red)[(a^3−b^3)]/ color(blue)[ (a^2−b^2)]#

#∙#Using the identities: #color(blue)(a^2-b^2=(a+b)(a-b)# and #color(red)(a^3-b^3= (a-b)(a^2+ab+b^2) #

#=> color(blue) [(a+b)(a-b)] /(a−b)-[color(red){(a-b)(a^2+ab+b^2)}]/color(blue)[(a+b)(a-b)]#

#=> [ (a+b) cancel ((a-b))] /cancel((a−b))-[cancel((a-b))(a^2+ab+b^2) ]/[(a+b)cancel((a-b))]#

#=> (a+b)-((a^2+ab+b^2))/((a+b)) #

#∙#Multiplying the numerator and denominator of the first term with #(a+b)#, since the LCM is #(a+b)#:

#=>(a+b)^2/((a+b))-((a^2+ab+b^2))/((a+b))#

#∙# Using the identity:#color(darkorange)((a+b)^2=(a^2+2ab+b^2)#

#=>color(darkorange)((a^2+2ab+b^2))/((a+b))-((a^2+ab+b^2))/((a+b))#

#=> [(a^2+2ab+b^2) - (a^2+ab+b^2)]/((a+b))#

#=> [ cancel a^2+2ab+cancelb^2- cancela^2-ab-cancelb^2]/((a+b))#

#color(magenta)(=>(ab)/(a+b)#

~Hope this helps! :)

Related questions
Impact of this question
1328 views around the world
You can reuse this answer
Creative Commons License