Binomial Expansion Help?

x = -788

x has to be less than 8/9 otherwise the expansion is nonsense

According to the mark scheme:

$[br]10\sqrt[3]{71} = 10\sqrt[3]{8-9x}[br]\therefore 71 = 8-9x[br][br]If x = 0.1;[br]71 = 8 - 9(0.1) = 7.1$
This is obviously impossible. Therefore I propose that the mark scheme is incorrect.

Hi I'm confused with part (b) to this question which is from the replacement June 2013 C4 paper. My expansion to part (a) is 2- 3/4x - 9/32x^2 - 45/256x^3

So for part (b) I did:

( 8 - 9x )^1/3 = ( 7100 )^1/3

8 - 9x = 7100

x = -788

And when I substituted this into the expansion, I got a really weird answer, which was wrong. The mark scheme says x=0.1

How do you go about answering this and why is my method wrong? Thank you

Yes, I agree there is a typo. However I think the value of x is correct. The error is in the cube root. It should read (7.1)^1/3 not 71.

So it's 7.1 as the modulus of x<8/9?

So it's 7.1 as the modulus of x<8/9?

To use your binomial expansion, you need a value of x with |x| < 8/9. Obviously you can't achieve this starting from 7100 itself, but 7100 = 1000 x 7.1 and you know what the cube root of 1000 is, so you work with 7.1 which gives you an x value of 0.1 i.e. small enough for the binomial expansion to be valid.

To use your binomial expansion, you need a value of x with |x| < 8/9. Obviously you can't achieve this starting from 7100 itself, but 7100 = 1000 x 7.1 and you know what the cube root of 1000 is, so you work with 7.1 which gives you an x value of 0.1 i.e. small enough for the binomial expansion to be valid.

I've just been stuck on this question (my C4 exam is tomorrow lol) and this explanation really helped, thanks!!!!!

Is this A2?