Factoring 64a³+8: got 8(8a³+1), but ans is 8(2a+1)(4a²-2a+1)

[fdragon]

Factoring 64a³+8: got 8(8a³+1), but ans is 8(2a+1)(4a²-2a+1)

The answer to this problem is given but I'm not sure how to get there.

I thought the answer was 8(8a³+1).... but the answer is 8(2a+1)(4a²-2a+1)

I can see how I can get from the final answer to 8(8a³+1) but I don't see how you get from 8(8a³+1) to the final answer.

How do you get to that answer?

 

[Deleted member 4993]

Re: Factoring 64a³+8

The answer to this problem is given but I'm not sure how to get there.

I thought the answer was 8(8a³+1).... but the answer is 8(2a+1)(4a²-2a+1)

I can see how I can get from the final answer to 8(8a³+1) but I don't see how you get from 8(8a³+1) to the final answer.

How do you get to that answer?

Hints:

\(\displaystyle 8 \, = \, 2^3\)

and

\(\displaystyle a^3 \, + \, b^3 \, = \, (a \, + \, b)\cdot(a^2 \, - \, ab \, + \, b^2)\)

 

[fdragon]

Re: Factoring 64a³+8

So is it always the case that the middle term in the second set of parentheses, which is 2a, is going to be the same as the first term in the other first of parentheses? ---> 8(2a+1)(4a²-2a+1)

 

[stapel]

Re: Factoring 64a³+8

So is it always the case that the middle term in the second set of parentheses, which is 2a, is going to be the same as the first term in the other first of parentheses?

No; as the formula (given in the previous reply) stated, the middle term of the three-term factor for a[sup:3cby3aim]3[/sup:3cby3aim] + b[sup:3cby3aim]3[/sup:3cby3aim] will always be "-ab". Naturally, if b = 1, then -ab will equal -a.

Eliz.