Dear Uncle Colin
I’m told that $x$ satisfies $x^3 - x-1=0$ , and I need to find the value of $\sqrt[3]{3x^2 - 4x} +x \sqrt[4]{2x^2 +3x + 2}$. Obviously, I could solve the cubic and plug the answer in, but I don’t think that’s in the spirit of it.
- Can Anyone Reduce Down Algebraic Nonsense?
Hi, CARDAN, and thanks for your message!
The first part drops out very nicely if you spot that the radicand1 is the difference between a binomial expansion and your equation: $(x^3 - x - 1) - (3x^2 - 4x) = x^3 - 3x^2 + 3x -1$, or $(x-1)^3$.
So $0 - (3x^2-4x) = (x-1)^3$, and $\sqrt[3]{3x^2-4x} = 1-x$. We’re off to a flyer!
Can we do something similar with the other root? Almost. I’m going to see if I can finagle it into the form of a fourth power.
First, I’m going to multiply in the $x$, which becomes $x^4$ under the square root: the radicand is now $2x^6 + 3x^5 + 2x^4$.
Now, $x^6 = x^2 + 2x + 1$, just squaring the constraint, and I can sub that in: the radicand becomes $3x^5 + 2x^4 + 2x^2 + 4x + 2$.
The $x^5$ is less straightforward; I’m going to make it $x^3 + x^2$, so the radicand is $2x^4 + 3x^3 + 5x^2 + 4x + 2$.
Ideally, I’m only going to have one $x^4$ under the root, so I’ll replace one of them with $x^2 + x$. Now I have $x^4 + 3x^3 + 6x^2 + 5x + 2$.
That’s excitingly close! I know that $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$. The difference between the two? $x^3 - x - 1$, which is zero!
So our radicand is $(x+1)^4$ and the second term is simply $x+1$.
The whole expression is $(1-x) + (x+1)$, or… drumroll… two.
Hope that helps!
- Uncle Colin
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the bit under the root ↩︎