A small help please, a big confusion x)

[Lexadis]

[We are currently studyin (a+b)³ = a³ + 3a²b + 3ab² + b³ ]
So this was the question:

Q. If (a+b) = 5 and ab = 6, evaluate a³+b³.

So, I tried this question in 2 different ways, both which I assumed correct, but I seem to get different answers x.X
Anyways, here it goes:
Method I
(a³+b³) = (a+b)³
= (a+b)(a+b)(a+b)
= 5 x 5 x 5
= 125.

Method II
But since they have also given the value of ab=6, I thought of using it too.So this was what I did:
(a³+b³) = (a+b)³
= (a+b)(a+b)²
= (a+b)(a²+2ab+b²)
= (a+b) (a²+b²+ 2ab)
= 5 (5² + 2x6)
= 5 (25 + 12)
= 5 x 37
= 185

???
Any idea as to what wrong I have done? Thank you very much c:

 

[Unknown008]

You have to start from the given expansion:

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

If (a+b) = 5 and ab = 6, evaluate \(\displaystyle a^3+b^3\).

This means:

\(\displaystyle (5)^3 = a^3 + 3a(6) + 3b(6) + b^3\)

Can you take it from here?

 

[soroban]

Hello, Lexadis!

This requires a clever trick.

We are currently studying: (a+b)³ = a³ + 3a²b + 3ab² + b³

Q. If (a+b) = 5 and ab = 6, evaluate a³+b³.

Since \(\displaystyle a+b \,=\,5\), then \(\displaystyle (a+b)^3 \:=\:5^3 \:=\:125\)

This means: .\(\displaystyle a^3 + 3a^2b+3ab^2+b^3 \:=\:125\)

Factor: .\(\displaystyle a^3 + 3(ab)(a+b) + b^3 \:=\:125\)
. . . . . . . . . . . . \(\displaystyle \uparrow \quad\;\;\;\uparrow\)
. . . . . . . . This is 6 . . This is 5

We have: .\(\displaystyle a^3 + 3(6)(5) + b^3 \:=\:125\)

. . . . . . . . . . . \(\displaystyle a^3 + 90 + b^3 \:=\:125\)

. . . . . . . . . . . . . . . \(\displaystyle a^3 + b^3 \:=\:35\)

 

[Lexadis]

Oh thank you both Unknown008 & soroban, you gusy have been a great help c:
So, I have another question of that sort, and I solved it using the method you all pointed out, and I just wanted to check whether it's right, any help? :]

Q. If \(\displaystyle (a-b) = 6\) and \(\displaystyle ab = 7, \)evaluate \(\displaystyle a^3-b^3\).
So, this was what I did:
---------------\(\displaystyle a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3\)
---------------\(\displaystyle a^3 - 3ab(a-b) - b^3 = 6^3\)
---------------\(\displaystyle a^3 - 3.7.6 - b^3= 216\)
---------------\(\displaystyle a^3 - 126 - b^3= 216\)
---------------\(\displaystyle a^3 - b^3= 216+126\)
---------------\(\displaystyle a^3-b^3= 342\)

Is that right? :]

 

[DrPhil]

Oh thank you both Unknown008 & soroban, you gusy have been a great help c:
So, I have another question of that sort, and I solved it using the method you all pointed out, and I just wanted to check whether it's right, any help? :]

Q. If \(\displaystyle (a-b) = 6\) and \(\displaystyle ab = 7, \)evaluate \(\displaystyle a^3-b^3\).
So, this was what I did:
---------------\(\displaystyle a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3\)
---------------\(\displaystyle a^3 - 3ab(a-b) - b^3 = 6^3\)
---------------\(\displaystyle a^3 - 3.7.6 - b^3= 216\)
---------------\(\displaystyle a^3 - 126 - b^3= 216\)
---------------\(\displaystyle a^3 - b^3= 216+126\)
---------------\(\displaystyle a^3-b^3= 342\)

Is that right? :]

Perfect!

 

[Lexadis]

Perfect!

Oh thank you