# Denesting Method for $\sqrt[3]{-10+9\sqrt{-3}}$ (Without Solving Another Cubic Equation)?

To solve the cubic equation $$x^3-2x^2-x+2=0$$such an answer is obtained: $$\sqrt[3]{-10+9\sqrt{-3}}$$ which can not be easily denested manually.

For denesting that radical, I had to solve another cubic equation. Sometimes, this second cubic equation would be more complex than the main equation. Is there any way for denesting a radical, which does not generate a new cubic equation? (hint: the denested form is $$\sqrt[3]{-10+9\sqrt{-3}}=2+\sqrt{-3}$$)

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• Well $x=2$ is a solution to the original cubic, so you're essentially solving a quadratic. Commented Aug 1 at 12:32
• thank you, but how you can find that? Commented Aug 1 at 12:47
• Rational root thereom. If on rational roots exist (which they might not) it must be $\pm 1,\pm 2$ as $x=2$ yield $x^2 = 2x^2$ and $x=2$ the sum is $0$. Commented Aug 1 at 15:43

$$\sqrt[3]{+9\sqrt{-3}-10}=\sqrt[3]{9i\sqrt{3}-10}$$

Using the hint you were given it is only natural to conclude that the original expression is $$\sqrt[3]{(2+\sqrt{-3})^3}$$.

We know $$(a+b)^3=a^3+3ab^2+3a^2b+b^3$$ in this case this becomes $$(2+i\sqrt{3})^3=a^3+3ab^2+3a^2b+b^3$$. The value of $$a$$ is the easiest to find since we already have the $$2$$ in the hinted expression. Then you can guess and check and it would not take long to see that $$a=2,b=i\sqrt{3}$$ satisfy the equality.

$$\sqrt[3]{(i\sqrt{3}+2)^3}=i\sqrt{3}+2=2+\sqrt{-3}$$

• But how did you come up with this expansion of $9i\sqrt{3}$? Commented Aug 1 at 12:34
• $\sqrt{-3}=\sqrt{-1}\times \sqrt{3}=i\sqrt{3}$ Commented Aug 1 at 12:35
• thank you, but how can we write the expression as what you said? Commented Aug 1 at 12:40
• $8-18=-10$ and $12i\sqrt{3}-3i\sqrt{3}=9i\sqrt{3}$ Commented Aug 1 at 12:41
• That's right. thank you dude. Commented Aug 1 at 12:49