# Hi guys I need help please!

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#1
I was doing a question but kept getting the answer wrong. On my final attempt my answer was correct but I'm not sure if the method was legitimate and that the maths was correct. Please can someone check for me? Here's the question but I've also attached it in writing so it's easier to see:

(8 a^3 b^4 / (2 a b^3)^2) ^3
MULTIPLIED BY
(3 a^8 b^3 / a^3 b)^4

Here's what I did:

I tackled the denominator of the first fraction, first.
(8 a^3 b^4 / 4 a^2 b^6) ^3
MULTIPLIED BY
(3 a^8 b^3 / a^3 b)^4

Then I did 8^3 , a^3*3, b^6*3 for the numerator of the first fraction and 4^3 , a^2*3, b^6*3 for the denominator of the first fraction
=(512 a^9 b^12 / 64 a^6 b^18)
MULTIPLIED BY
Then I did the same for the other fraction, so for the numerator: 3^4 , a^8*4 , b^3*4 and a^3*4 , b^1*4 for the denominator
=(81 a^32 b^12 / a^12 b^4)

So if we bring these both together it is the following:
(512 a^9 b^12 / 64 a^6 b^18)
MULTIPLED BY
(81 a^32 b^12 / a^12 b^4)

I multiplied across, so I did the numerator of fraction 1 * numerator of fraction 2, and denominator of fraction 1* denominator of fraction 2
=41472 a^41 b^24 / 64 a^18 b^22

Finally I divided the numerator by the denominator, so

=648 a^23 b^2

This answer is correct. Is my working correct though or was I just lucky?

Thanks!
0
1 year ago
#2
Your method is mathematically correct so it wasn't just luck that you got it right.
You would have found it easier (smaller numbers, less arithmetic) if you had cancelled down fractions at every possible opportunity. For example (8 a^3 b^4 / 4 a^2 b^6) ^3 cancels to (2a/b²)³.
2
#3
Thanks so much, Mark! I appreciate that. I didn't even realise at that time that I could or should simplify it further
0
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