need help to find 3rd square root 8x^3-280 = 2x-10

[amyjo24]

I'm trying to figure out the answer to 3rd square root 8x^3-280 = 2x-10.
in order to get rid of the square root i understand you need to cube both sides. i am getting confused on cubing 2x-10. i think the easiest way is to do (2x-10)(2x-10) then that answer times (2x-10) could some help me please in writing out all the steps for me to understand. thanks

 

[Loren]

Re: need help

This might make it a little easier for you. Yes. To cube a binomial such as a-b, you need to multiply (a-b)(a-b) getting a[sup:jat9ur8u]2[/sup:jat9ur8u]-2ab+b[sup:jat9ur8u]2[/sup:jat9ur8u], then multiply that by a-b.
\(\displaystyle \sqrt[3]{8x^3-280} = 2x-10\)
\(\displaystyle \sqrt[3]{8(x^3-35)} = 2(x-5)\)
Can you figure out a way to divide both sides by the same number to make your equation simpler?

 

[Deleted member 4993]

Re: need help

This might make it a little easier for you. Yes. To cube a binomial such as a-b, you need to multiply (a-b)(a-b) getting a[sup:s83oc4lp]2[/sup:s83oc4lp]-2ab+b[sup:s83oc4lp]2[/sup:s83oc4lp], then multiply that by a-b.
\(\displaystyle \sqrt[3]{8x^3-280} = 2x-10\)
\(\displaystyle \sqrt[3]{8(x^3-35)} = 2(x-5)\)
Can you figure out a way to divide both sides by the same number to make your equation simpler?

Or - take cube of both sides first:

\(\displaystyle \sqrt[3]{8(x^3-35)} = 2(x-5)\)

\(\displaystyle [\sqrt[3]{8(x^3-35)}]^3 \, = \, [2(x-5)]^3\)

\(\displaystyle 8(x^3-35) \, = \, 8(x-5)^3\)

\(\displaystyle x^3 - 35 \, = \, x^3 - 15x^2 + 75x - 125\)

Now continue....