Question 166513
For x²(a-b)+ a²(b-x)+ b²(x-a) show that (x-a) is a linear factor.
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Multiply to eliminate the brackets, we can leave the last term alone
x^2a - x^2b + a^2b - xa^2 + b^2(x-a)
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Regroup:
x^2a - xa^2 + a^2b - x^2b + b^2(x-a)
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Factor out xa and b
xa(x-a) + b(a^2 - x^2) + b^2(x-a)
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Change the sign of b  to -, to change signs inside the brackets
xa(x-a) - b(x^2 - b^2) + b^2(x-a)
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Note that x^2 - b^2 is the "difference of squares and can be factored
xa(x-a) - b(x-a)(x+a) + b^2(x-a)
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Note that each term contains the factor (x-a) now