MHB Finding a^-3 + b^-3; Can someone check if what I did is right?

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To find the value of a^-3 + b^-3 given that ab = 1 and a^2 + b^2 = 4, it is established that a and b are reciprocals. The expression a^-3 + b^-3 can be rewritten as a^3 + b^3, which is calculated using the formula (a+b)(a^2 - ab + b^2). Substituting the known values, a^3 + b^3 simplifies to (a+b)(3). To find a + b, the equation a^2 + 2ab + b^2 = 6 leads to a + b = √6. The final result for a^-3 + b^-3 is then √6 multiplied by 3.
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Can someone check if this is right?

So, having two numbers, a and b, we can say that their product is 1, and the sum of their squares is 4, find the sum of:

$$a^{-3} + b^{-3}$$

Well, we have:
$$ab = 1$$

$$a^2 + b^2 = 4$$

This means that a and b are reciprocals... Thus:

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

$$a^3 + b^3 = (a+b) (a^2 - ab + b^2)$$

We know that a^2 +b^2 = 4 and that ab = 1, so we put those in, respectively.

$$a^3 + b^3 = (a+b) (4 - 1)$$

$$a^3 + b^3 = (a+b) (3)$$

And now the problem is finding a + b, which isn't that hard either.

We can take ab = 1 and multiply it by two and add a^2 + b^2 to both sides, which is 4.

$$a^2 + 2ab + b^2= 2 + 4 = 6$$

We factorize, then square root it.

$$\sqrt{(a+b)^2} = \sqrt{6} ; a+b = \sqrt{6}$$

Then, we just insert that into what we previously had:

$$a^3 + b^3 = \sqrt{6} (3)$$

Is that right?
 
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I agree.
 
What if $a,b < 0$?
 
Deveno said:
What if $a,b < 0$?

My bad, I didn't include that: The COMPLETE instructions are:

Let a and b be real, positive numbers such that their product is one and the sum of their squares is 4. Find the exact value of the expression:

$$a^{-3} + b^{-3}$$
 
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