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Simplify the expression. 7x to the -3power times 6x to the 3power (sorry dont know how to make it look like an actual problem)
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7(x)^(-3) * 6(x)^3 I believe?
Yeah that makes more sense, thanks
Okay, you know how to solve x^3-y^3 type of equations?
yeah, I know that i would multiply 6 and7 which =42 but i dont know what to do with the -3 and 3
Take x to be cuberoot(7)*(x)^(-1) and y to be cuberoot(6)*x^1
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would it end up being \[\frac{ 42 }{ x^5 }\] ?
No. the xs get cancelled out.
The exponents have a property a^m . a^n = a^(m+n) okay? Now, a here is x m is 3, n is -3
Clear, I believe?
i think i got it
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So, what really happened was, (7)(6)x^(+3-3) Which is (42)(x)^0 And that is 42*1 which is 42. Clear?
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