Factoring Problem

[TangoFoxtrotGolf]

Factor 8 - (a + 1)^3

I started cubing the (a + 1) binomial, resulting in

8 - (a^3 + 3a^2 + 3a + 1), and then subtracting to get

8 - a^3 - 3a^2 - 3a - 1, and then subtracting 1 from 8 I get,

7 - a^3 - 3a^2 -3a

I do not see what to do next.

 

[Unco]

Hi Tango,

Recall the identity: \(\displaystyle x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)\).

An acronym due to Soroban helps us to remember the signs in the right-hand side: S.O.A.P. for Same-Opposite-Always Positive.

So, returning to the initial expression, 8 - (a + 1)^3, your first task is to choose appropriate values for x and y in the identity above.

 

[Loren]

Factor 8 - (a + 1)^3

I started cubing the (a + 1) binomial, resulting in

8 - (a^3 + 3a^2 + 3a + 1), and then subtracting to get

8 - a^3 - 3a^2 - 3a - 1, and then subtracting 1 from 8 I get,

7 - a^3 - 3a^2 -3a

I do not see what to do next.

Hint: 2[sup:thakkwuk]3[/sup:thakkwuk] - b[sup:thakkwuk]3[/sup:thakkwuk] = (2-b)(2[sup:thakkwuk]2[/sup:thakkwuk]+2b+b[sup:thakkwuk]2[/sup:thakkwuk])

 

[TangoFoxtrotGolf]

Thanks Unco,

I got it,

x = 2
y = a + 1

And then it works out perfectly as a difference of two cubes.

Thanks,

 

[TangoFoxtrotGolf]

Thanks Loren,

Read Unco's reply first, but your answer was essentially the same.

And, I get the difference of two cubes now.

Thanks,

 

[mmm4444bot]

… An acronym due to Soroban helps us to remember the signs in the right-hand side: S.O.A.P. for Same-Opposite-Always Positive.

I like. Thanks for that.