IHateFactorial
- 15
- 0
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?
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?