factorize \( m^{3}+n^{3} \) \( a^{12}-b^{6} \) \( 8 x^{3}-y^{9} \) \( x^{3}+8 y^{3} \)
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To factor \(m^3 + n^3\), you can use the sum of cubes formula: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). This gives us \(m^3 + n^3 = (m + n)(m^2 - mn + n^2)\). Next for \(a^{12} - b^6\), this can be viewed as the difference of squares: \(a^{12} - (b^3)^2\). Using the difference of squares formula, we have: \((a^6 - b^3)(a^6 + b^3)\). Further factorization of \(a^6 - b^3\) and \(a^6 + b^3\) can be applied subsequently. For \(8x^3 - y^9\), factor out the greatest common factor if possible. Notice that \(8x^3\) can be rewritten as \((2x)^3\) and \(y^9\) as \((y^3)^3\), allowing you to express it as a difference of cubes: \((2x)^3 - (y^3)^3 = (2x - y^3)((2x)^2 + 2xy^3 + (y^3)^2)\). Lastly, \(x^3 + 8y^3\) can similarly be factored using the sum of cubes formula. Here, \(8y^3\) is recognized as \((2y)^3\), thus yielding \(x^3 + (2y)^3 = (x + 2y)(x^2 - 2xy + 4y^2)\). So, the results are: - \(m^3 + n^3 = (m + n)(m^2 - mn + n^2)\) - \(a^{12} - b^6 = (a^6 - b^3)(a^6 + b^3)\) - \(8x^3 - y^9 = (2x - y^3)((2x)^2 + 2xy^3 + (y^3)^2)\) - \(x^3 + 8y^3 = (x + 2y)(x^2 - 2xy + 4y^2)\)