Cubic Identities (1 Viewer)

Buwwy · Feb 8, 2012

I know the answer but I just can't seem to get it.
Please help!

Question is found in Yr 11 Cambridge Maths 3U book, Exercise 1E, Q16 (pg 12)

Simplify (1+a)^2/[1+(a/(1-a+(a/(1+a+a^2)))]

This is the image of the question.

Any help is appreciated. Thanks

Shadowdude · Feb 8, 2012

$\frac{(1+a)^{2}}{1+\frac{a}{1-a+\frac{a}{1+a+a^{2}}}}$

Adding the two bottom-most fractions:

$=\frac{(1+a)^{2}}{1+\frac{a}{\frac{(1-a)(1+a+a^{2})+a}{1+a+a^{2}}}}$

Simplifying:

$=\frac{(1+a)^{2}}{1+\frac{a}{\left ( \frac{(1-a^{3})+a}{1+a+a^{2}}} \right )}$

Rearranging the bottom fraction:

$= \frac{(1+a)^{2}}{1+\frac{a(1+a+a^{2})}{1-a^{3}+a}}$

Adding the fractions:

$= \frac{(1+a)^{2}}{\frac{1-a^{3}+a+(a(1+a+a^{2}))}{1-a^{3}+a}}$

Rearranging the fraction:

$= \frac{(1+a)^{2}(1-a^{3}+a)}{1-a^{3}+a+a+a^{2}+a^{3}}$

Simplifying:

$= \frac{(1+a)^{2}(1-a^{3}+a)}{1+2a+a^{2}}$

The bottom simplifies to:

$= \frac{(1+a)^{2}(1-a^{3}+a)}{(1+a)^{2}}$

Cancelling:

$= 1 - a^{3} +a$

Timske · Feb 8, 2012

cambridge has retarded questions

Carrotsticks · Feb 12, 2012

Timske said:
cambridge has retarded questions

No, they have beautiful questions.

math man · Feb 12, 2012

Carrotsticks said:
No, they have beautiful questions.

This

Cubic Identities (1 Viewer)

Buwwy

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Shadowdude

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Timske

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Carrotsticks

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math man

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