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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

How to solve this algebra problem without a calculator?

The answer's 2.

Original post by Kurraiyo

The answer's 2.

\left(a-\frac{1}{a}\right)^2=...

\left(a-\frac{1}{a}\right)^3=...

Original post by Kurraiyo

The answer's 2.

The second power of a+1/a or a-1/a differ from a^2+1/a^2 or a^21/a^2 in a constant only

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

So

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

Similarly as

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

so

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

I can feel one of my heads coming on...

OP

Original post by ztibor

The second power of a+1/a or a-1/a differ from a^2+1/a^2 or a^21/a^2 in a constant only

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

So

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

Similarly as

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

so

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

Original post by BabyMaths

\left(a-\frac{1}{a}\right)^2=...

\left(a-\frac{1}{a}\right)^3=...

Thank you!! :smile:

:smile:

OP

Bump, another similar question I'm stuck on.

Original post by Kurraiyo

Bump, another similar question I'm stuck on.

You can expand

(\omega+1)^5=(-\omega^2)^5

.

The left side simplifies nicely using the fact that

\omega^2+\omega+1=0

.

OP

Original post by BabyMaths

You can expand

(\omega+1)^5=(-\omega^2)^5

.

The left side simplifies nicely using the fact that

\omega^2+\omega+1=0

.

Thanks again you're a lifesaver :smile:

:smile:

Original post by Kurraiyo

Thanks again you're a lifesaver :smile:

:smile:

Steady on there. :tongue:

:tongue:

:smile:

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