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[sin(x)^3 + cos(x)^3] / [sin(x) + cos(x)]

= [(sin(x) + cos(x))(sin(x)^2 - sin(x)cos(x) + cos(x)^2)] / [sin(x) + cos(x)]

***Now you can cancel a "sin(x) + cos(x)" from the top and bottom of the fraction. This makes the bottom of the fraction equal to 1. I am just going to write the next step without a 1 on the bottom of the fraction (x/1=x).

So now you just have:

= (sin(x)^2 - sin(x)cos(x) + cos(x)^2) *I'm going to move some terms around now. ~

Not doing any computation in this step.

= (sin(x)^2 + cos(x)^2 - sin(x)cos(x)) *Now we know that cos(x)^2 + sin(x)^2 = 1.

= 1 - sin(x)cos(x)

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13y ago

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