Help with negative exponents please: (c^-1 + d^-1) \ (c^-2 + d^-2)
hint: multiply every term by c^2d^2
I did that and got cd (d + c) \ (d^2 + c^2)...can this be simplified?
I'm getting \[\large \frac{cd(d+c)}{d^2 + c^2}\] which is what you have and you can't simplify further so nice work
thank you so much! are you available to help with another problem?
sure I can do one more
thank you :) (x^-3 + y^-3) \ (x^-1 + y^-1)
multiply everything by x^3y^3 and tell me what you get
(y^3 + x^3) \ x^2y^3
multiply each term by x^3y^3 to get x^-3 times x^3y^3 = y^3 y^-3 times x^3y^3 = x^3 so that part is correct in the numerator
in the denominator we have x^-1 times x^3y^3 = x^2y^3 y^-1 times x^3y^3 = x^3y^2
so the final answer is \[\large \frac{y^3+x^3}{x^2y^3+x^3y^2}\] and optionally you can factor to get \[\large \frac{y^3+x^3}{x^2y^2(y+x)}\]
hi @jim_thompson5910 did u get my PM?
oh I guess you can factor the numerator as well to get y^3 + x^3 = (y+x)(y^2 + yx + x^2)
how do you get the + x^3y^2 in the denominator?
oh i see that in the numerator... would that be the same as (x + y) (x^2 + xy + y^2)?
in the denominator, we have the terms x^-1 + y^-1 multiply those term by x^3y^3 to get x^2y^3+x^3y^2
sorry I see
sorry made a typo, it should be y^3 + x^3 = (y+x)(y^2 - yx + x^2)
so the full step by step picture looks like this
\[\large \frac{x^{-3} + y^{-3}}{x^{-1} + y^{-1}}\] \[\large \frac{x^3y^3*x^{-3} + x^3y^3*y^{-3}}{x^3y^3*x^{-1} + x^3y^3*y^{-1}}\] \[\large \frac{y^3+x^3}{x^2y^3+x^3y^2}\] \[\large \frac{(y+x)(y^2 - yx + x^2)}{x^2y^2(y+x)}\] \[\large \frac{\cancel{(y+x)}(y^2 - yx + x^2)}{x^2y^2\cancel{(y+x)}}\] \[\large \frac{y^2 - yx + x^2}{x^2y^2}\] ------------------------------------------------------------------------------------------ Final Answer: \[\large \frac{y^2 - yx + x^2}{x^2y^2}\]
thank you so much for taking the time to explain it so well!
yw
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