Simplification help!

[Juliannimal]

Simplify with no fraction in the numerator or denominator.

(2-x[sup:2jd88czr]3[/sup:2jd88czr])[sup:2jd88czr]-1/2[/sup:2jd88czr](-x[sup:2jd88czr]3[/sup:2jd88czr])(3-x[sup:2jd88czr]3[/sup:2jd88czr])[sup:2jd88czr]2[/sup:2jd88czr]-(2-x[sup:2jd88czr]3[/sup:2jd88czr])[sup:2jd88czr]1/2[/sup:2jd88czr](3-x[sup:2jd88czr]3[/sup:2jd88czr])(1/2x[sup:2jd88czr]2[/sup:2jd88czr])


The answer is:
x[sup:2jd88czr]2[/sup:2jd88czr](3-x[sup:2jd88czr]3[/sup:2jd88czr])(3x[sup:2jd88czr]3[/sup:2jd88czr]-8) / 2 square root 2-x[sup:2jd88czr]3[/sup:2jd88czr]

I don't know how to get this answer!! Please Help!

 

[Deleted member 4993]

Simplify with no fraction in the numerator or denominator.

(2-x[sup:qtc1v3sy]3[/sup:qtc1v3sy])[sup:qtc1v3sy]-1/2[/sup:qtc1v3sy](-x[sup:qtc1v3sy]3[/sup:qtc1v3sy])(3-x[sup:qtc1v3sy]3[/sup:qtc1v3sy])[sup:qtc1v3sy]2[/sup:qtc1v3sy]-(2-x[sup:qtc1v3sy]3[/sup:qtc1v3sy])[sup:qtc1v3sy]1/2[/sup:qtc1v3sy](3-x[sup:qtc1v3sy]3[/sup:qtc1v3sy])(1/2x[sup:qtc1v3sy]2[/sup:qtc1v3sy])


The answer is:
x[sup:qtc1v3sy]2[/sup:qtc1v3sy](3-x[sup:qtc1v3sy]3[/sup:qtc1v3sy])(3x[sup:qtc1v3sy]3[/sup:qtc1v3sy]-8) / 2 square root 2-x[sup:qtc1v3sy]3[/sup:qtc1v3sy]

I don't know how to get this answer!! Please Help!]

\(\displaystyle \frac{1}{\sqrt{2-x^3}} \cdot (-x^3)\cdot (3-x^3)^2 - \sqrt{2-x^3}\cdot (3-x^3)\cdot \frac{1}{2}\cdot x^2\)

This is what you have posted (equivalent) - is that correct problem?

Also, please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[Juliannimal]

I understand how to get the denominator but not how to get the numerator. And yes, that is the equivalent.

 

[chrisr]

I get \(\displaystyle (3-x^3)(x^2)(2x^4+x^3-6x-2)\) for the numerator.

On the right-hand side, you can write \(\displaystyle \sqrt{2-x^3}\) as \(\displaystyle \frac{2-x^3}{\sqrt{2-x^3}}\)

 

[soroban]

Hello, Juliannimal!

There is a typo . . .

. . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \downarrow\)
\(\displaystyle \text{Simplify: }\2-x^3)^{-\frac{1}{2}}(-x^2)(3-x^3)^2 \:-\2-x^3^{\frac{1}{2}}(3-x^3)\left(\tfrac{1}{2}x^2\right)\)

\(\displaystyle \text{The answer is: }\frac{x^2(3-x^3)(3x^3 - 8)} {2\sqrt{2-x^3}}\)

\(\displaystyle \text{We have: }\;-x^2(2-x^3)^{-\frac{1}{2}}(3-x^3)^2 \:-\:\tfrac{1}{2}x^2(2-x^3)^{\frac{1}{2}}(3-x^3)\)

\(\displaystyle \text{Factor: }\;\tfrac{1}{2}x^2(2-x^3)^{-\frac{1}{2}}(3-x^3)\,\bigg[-2(3-x^3) - (2-x^3)\bigg]\)

. . . . \(\displaystyle =\;\frac{x^2(3-x^3)} {2(2-x^3)^{\frac{1}{2}}}\,\bigg[-6 + 2x^3 - 2 + x^3\bigg]\)

. . . . \(\displaystyle =\; \frac{x^2(3-x^3)(3x^3-8)}{\sqrt{2-x^3}}\)

 

[chrisr]

Juliannimal !

great name !