Cancellation problem?

Question

how does [b/(a - b)] x [(a^3 - b^3)/ b]

cancel to (a^2 + b^2 + ab) ?

Please show me the steps involved thanks

Its the last step in the derivation of a frustum

Answer 1

[b/(a-b)] x [(a^3 - b^3)/b]
first the the numerator of the first term and the denominator of the second term cancel each other, thats pretty straight forward.
so we have (a^3 - b^3 )/ (a-b) since we multiplied the two terms.
but we can factor a^3 - b^3 to (a-b)(a^2 + b^2 + ab) (pecfect cube)
so we get (a-b)(a^2 +b^2 + ab)/(a-b)
but (a-b) can be cancel out
so we get a^2 + b^2 + ab

there u go dude
pick me as ur best answer, lol

Answer 2

so you have the equation as you can see the b's cancel leaving you [1/(a-b)]x [a^3-b^3] then you factor the a^3-b^3 which is
(a-b)(a^2 + ab +b^2) then the (a-b) cancels and your left with

a^2 + ab + b^2



side note when you factor cubes like a^3 -b^3 and others you take the cube root like for instance (a -b) then you square a and b then you take the product of a and b and switch the signs


hope this helped if not message me i can explain better

Answer 3

what exact math is this? There is probally a easy solution but im just fried today so i cant some up with the answer..but heres what i got..

[b/a+b X a^3-b^3/b]

= a^3b-b^4/ ab-b^2
= a^3-b^3/a-b

Not sure where else to go...sorry

Answer 4

Because (a^3 -b^3) = (a-b)(a^2 +ab +b^2)]
[b/(a - b)] x [(a^3 - b^3)/ b]
={b/(a-b)}x[(a-b)(a^2 +ab +b^2)]/b
cancel b(a-b) from numerator and denominator
=(a^2 +ab +b^2)]

Answer 5

[b/(a - b)] x [(a³ - b³)/ b] = [b/(a - b)] x [(a -b)(a² + ab + b²)]/b so you can cancel the "b" on the top with the "b" which is in the bottom. and you can do it for (a - b) as well. finally, what is left is (a² + 2ab + b²).

Note: you know that you can simplify (a³ - b³) to (a - b)x(a² + 2ab +b²)

Answer 6

[b/(a - b)] x [(a^3 - b^3)/ b]
= (a^3 - b^3)/(a-b)
= (a-b)(a^2 + ab + b^2)/(a-b)
= (a^2 + b^2 + ab)


NB: You have to first factorise (a^3 - b^3).

Answer 7

not telling you, thats cheating!