Special Factoring Problem...?
How would 8a^3 - 27b^6 factor? I've been looking through some of my old pre-calculus notes, but can't discover how to factor these "special" polynomials. Thank you.
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How would 8a^3 - 27b^6 factor? I've been looking through some of my old pre-calculus notes, but can't discover how to factor these "special" polynomials. Thank you.
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8a^3-27b^6
=(2a)^3-(3b^2)^3
=(2a-3b^2)[(2a)^2+
(3b^2)^2+(2a)*(3b^2)]
=(2a-3b^2)(4a^2+9b^4+6ab^2)
Take the optimum undemanding element 3 out. you have 3(4x^2-12x+9) permit 3(4x^2-12x+9) = 0 subsequently 4x^2-12x+9=0 via factoring, you will get (2x-3)^2. subsequently, the respond would be 3(2x-3)^2