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Choose one of the factors of 24x6 – 1029y3 3 2x2 – 7y 4x4 + 14x2y + 49y2 All of the above
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\(\Large24x^6-1029y^3=3*2^3(x^2)^3-3*(3^2)^3y^3\) \(\Large=3*(2x^2)^3-3(3^2y)^3\) \(\Large=3*(2x^2)^2-3*(9y)^3\) \(\Large(a^3-b^3)=(a-b)(a^2+ab+b^2)\) Does that help?
one number finishes in 9 so the only factors will be 3 and 9, but 9 is not a factor of 24, so you're only left with 3 . since no variables are common to both . the common factor is only 3
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