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Question

Factorise

(a-2)^3 - (a+1)^3

Show all working out and answer.

Thanks

Answer 1

(a - 2)^3 - (a + 1)^3

This is a difference of cubes, so we factor it like so.

[(a - 2) - (a + 1)] [ (a - 2)^2 + (a - 2)(a + 1) + (a + 1)^2 ]

Simplifying,

[-3] [ a^2 - 4a + 4 + a^2 - a - 2 + a^2 + 2a + 1 ]

(-3) [3a^2 - 3a + 3]
(-3)(3)(a^2 - a + 1)
(-9)(a^2 - a + 1)

Answer 2

(a - 2)3 - (a + 1)3

= [a3 - 3.2a2 + 3.2.2a - 2.2.2] - [a3 + 3a2 + 3a + 1]

= (a3 - a3) - (6a2 + 3a2) + (12a - 3a) - (8 + 1)

= 0 - 9a2 + 9a - 9

= -9(a2 - a + 1)

Hmm. The superscripts didn't quite come across when I did a copy & paste. I hope, however, that you get the idea.

Answer 3

(a-2)^3 - (a+1)^3
is the same as

[(a-2)x(a-2)x(a-2)] - [(a+1)x(a+1)x(a+1)]

mulitiply

[(a^2-4a+4)x(a-2)] - [(a^2+2a+1)x(a+1)]
[a^3-4a^2+4a-2a^2+8a-8] - [a^3+2a^2+a+a^2+2a+1]
[a^3-6a^2+12a-8] - [a^3+3a^2+3a+1]

don't forget we has to change all the signs because you have a minus sign

a^3-6a^2+12a-8 - a^3-3a^2-3a-1

now we combine like terms

-9a^2+9a-9

you can pull out (-9) from the entire problem

-9[a^2-a+1]

and voila, we're done!hope this helps!!!!

Answer 4

I gave the formula in the previous question. You just have to apply it.

Answer 5

a^3-2^3-3*a^2*2+3*a*2^2-a^3-1^3-3*a^2*1-3*a*1^2=
a^3-8-6*a^2+12*a-a^3-3-3a^2-3a=
3a^2+9a-11

Answer 6

same as puggy.