I need urgent help with rational expressions?

Question

1) Reduce [ (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 ] / [ (a^4 - b^4)^3 + (b^4 - c^4)^3 + (c^4 - a^4)^3]


2) Solve:
[(x^3 - 1)/x + 4][(2x - 7)/(x^2 + 3x + 1)]

There are two questions to solve. 1) and 2) are two separate questions. You must simplify the expression in the first question and solve the second question

Answer 1

I don't know what you mean by 2), but I'll help you with 1):

First note that if x+y+z = 0, then x^3+y^3+z^3 = 3xyz. Since this applies to both numerator and denominator, we find they are actually products, not sums. Then using the difference of two squares rule, you should find the denominator is the product of three factors, and the numerator is 1.

Steve

EDIT - it is not possible to "solve" an expression, so still, I don't know what you want.

Answer 2

if( a+b+c) = 0 then a^3+b^3+c^3 = 3abc

so (a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2) = 3(a^2-b^2)(b^2-c^2)(c^2-a^2)

so (a^4-b^4)^3 + (b^4-c^4)^3 + (c^4-a^4) = 3(a^4-b^4)(b^4-c42)(c^4-a^4)

dividing we get ((a^2-b^2)(a^4-b^4))(b^2-c^2)/(b^4-c^4)))(c^2-a^2)/(c^4-a^4))

= 1/((a^2+b^2)(b^2+c^2)(c^2+a^2))

as (a^2-b^2)/(a^4-b^4) = 1/(a^2+b^2)

2)

(x^3-1)/(x+4) = (x-1)(x^2+x+1)

so we get expression = (x-1)(2x-7)/(x+4)

Answer 3

Try using these free algebra calculators