Troubleshooting Inverse Function Problem | Basic Algebra Review

[aquitaine]

Ok, I decided to review basic algebra since I haven't done anything with it in like, forever. I came across an inverse function problem that I can't get the right answer.

the equation is:

y = cuberoot(x+sqrt(1+x^2)) + cuberoot(x-sqrt(1+x^2))

I tried replacing X with Y, and solving for Y
and getting rid of the cube roots by cubing both sides
X^3 = y + sqrt(1+y^2) + y - sqrt(1+y^2)
simplifying a bit (the square roots go away)
x^3 = 2y
so
y = (1/2)x^3

Yet the book I'm using says the answer is y=(1/2)(3x+x^3)

What did I do wrong?

 

[HallsofIvy]

(a+ b)³ is NOT a³+ b³

It is a³+ 3a²b+ 3ab²+ b³

 

[djeitnstine]

Ok, I decided to review basic algebra since I haven't done anything with it in like, forever. I came across an inverse function problem that I can't get the right answer.

the equation is:

y = cuberoot(x+sqrt(1+x^2)) + cuberoot(x-sqrt(1+x^2))

I tried replacing X with Y, and solving for Y
and getting rid of the cube roots by cubing both sides
X^3 = y + sqrt(1+y^2) + y - sqrt(1+y^2)
simplifying a bit (the square roots go away)
x^3 = 2y
so
y = (1/2)x^3

Yet the book I'm using says the answer is y=(1/2)(3x+x^3)

What did I do wrong?

Your equation is a tad confusing, you have y = $$\sqrt[3]{x+sqrt(1+x^2)}$$ + $$\sqrt[3]{x-sqrt(1+x^2)}$$

then you cube both sides and switch y's and x's to get $$x^{3}$$ = y+$$\sqrt(1+y^2)$$ + y-$$\sqrt(1+y^2)$$

Which is wrong. Now if your equation was y = $$\sqrt[3]{x+\sqrt(1+x^2)+x-\sqrt(1+x^2)}$$ this method would be correct. However, it would have been simplified easily before you even cube both sides; In this case it would be in the form y = $$\sqrt[3]{2x}$$ for the roots cancel automatically.