Multiply and simplify

Question

Multiply and simplify. Assume that all variables in a radicand represent positive real numbers.

 

³√20x¹⁰y × ³√16x¹⁶y¹³

 

 

 

 

 

 

 

( answer is 4x⁸y^4 3√5x²y² but i need help step by step on how they got this answer)

Answer 1

Okay, the first thing you want to do is combine your two cube roots.  You do this by multiplying the pieces together.  Remember, if you have x³*x⁷, you have 10 different x's to multiply together.  You are adding the exponents.

 

That said, you now have the third root of (20*16*x²⁶y¹⁴)

From here, you want to remove numbers in groups of 3.  Looking at the numerical part first, you have 20*16, or 2²*5*2⁴, which can be combined to get 2⁶*5.

 

This is 2 groups of 3 for the 6's, so you can pull that totally out of the radical as 2² or 4.  The 5 is stuck inside.

 

Looking at x²⁶, how many groups of THREE x's can you get?  8*3=24, so 8 groups, with 2 leftover.  This is the x⁸ on the outside, with the x² left inside.

 

Looking at y¹⁴ you have 4*3=12, so 4 y's come out (y⁴) with again 2 y's left over for y² inside.

Answer 2

(³√20x¹⁰y) × (³√16x¹⁶y¹³)

 

To simplify, we want to factor the radicands (terms under the radical symbols) to find as many perfect cubes as we can.  These are the perfect cubes:

• x¹⁰ = (x³)³*x
• 16 = 8*2 = 2³*2
• x¹⁶ = (x⁵)³*x
• y¹³ = (y⁴)³y

 

Let's plug them in:

 

= (³√20(x³)³xy) × (³√2³2(x⁵)³x(y⁴)³y)

 

= (x^{3 3}√20xy) × (2x⁵y^{4 3}√2xy)

 

Now multiply the terms together, coefficients x coefficients and radicals x radicals:

 

= 2x⁸y^{4 3}√40x²y²

 

Now 40 = 8*5 = 2³*5, so:

 

= 2x⁸y^{4 3}√2³5x²y²

 

= 4x⁸y^{4 3}√5x²y²