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factorize \( m^{3}+n^{3} \) \( a^{12}-b^{6} \) \( 8 x^{3}-y^{9} \) \( x^{3}+8 y^{3} \)

Ask by Pollard Morrison. in South Africa
Feb 21,2025

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Answer

\( m^{3}+n^{3} = (m+n)(m^{2}-mn+n^{2}) \) \( a^{12}-b^{6} = (a^{2}-b)(a^{4}+a^{2}b+b^{2})(a^{2}+b)(a^{4}-a^{2}b+b^{2}) \) \( 8 x^{3}-y^{9} = (2x-y^{3})(4x^{2}+2xy^{3}+y^{6}) \) \( x^{3}+8 y^{3} = (x+2y)(x^{2}-2xy+4y^{2}) \)

Solution

Factor the expression \( m^{3}+n^{3} \). Factor the expression by following steps: - step0: Factor: \(m^{3}+n^{3}\) - step1: Factor the expression: \(\left(m+n\right)\left(m^{2}-mn+n^{2}\right)\) Factor the expression \( a^{12}-b^{6} \). Factor the expression by following steps: - step0: Factor: \(a^{12}-b^{6}\) - step1: Rewrite the expression: \(\left(a^{6}\right)^{2}-\left(b^{3}\right)^{2}\) - step2: Factor the expression: \(\left(a^{6}-b^{3}\right)\left(a^{6}+b^{3}\right)\) - step3: Factor the expression: \(\left(a^{2}-b\right)\left(a^{4}+a^{2}b+b^{2}\right)\left(a^{6}+b^{3}\right)\) - step4: Factor the expression: \(\left(a^{2}-b\right)\left(a^{4}+a^{2}b+b^{2}\right)\left(a^{2}+b\right)\left(a^{4}-a^{2}b+b^{2}\right)\) Factor the expression \( 8 x^{3}-y^{9} \). Factor the expression by following steps: - step0: Factor: \(8x^{3}-y^{9}\) - step1: Rewrite the expression: \(\left(2x\right)^{3}-y^{9}\) - step2: Factor the expression: \(\left(2x-y^{3}\right)\left(\left(2x\right)^{2}+2xy^{3}+y^{6}\right)\) - step3: Evaluate: \(\left(2x-y^{3}\right)\left(4x^{2}+2xy^{3}+y^{6}\right)\) Factor the expression \( x^{3}+8 y^{3} \). Factor the expression by following steps: - step0: Factor: \(x^{3}+8y^{3}\) - step1: Rewrite the expression: \(x^{3}+\left(2y\right)^{3}\) - step2: Factor the expression: \(\left(x+2y\right)\left(x^{2}-x\times 2y+\left(2y\right)^{2}\right)\) - step3: Calculate: \(\left(x+2y\right)\left(x^{2}-2xy+4y^{2}\right)\) The factorization of the given expressions is as follows: 1. \( m^{3}+n^{3} = (m+n)(m^{2}-mn+n^{2}) \) 2. \( a^{12}-b^{6} = (a^{2}-b)(a^{4}+a^{2}b+b^{2})(a^{2}+b)(a^{4}-a^{2}b+b^{2}) \) 3. \( 8 x^{3}-y^{9} = (2x-y^{3})(4x^{2}+2xy^{3}+y^{6}) \) 4. \( x^{3}+8 y^{3} = (x+2y)(x^{2}-2xy+4y^{2}) \)

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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: - \(m^3 + n^3 = (m + n)(m^2 - mn + n^2)\) - \(a^{12} - b^6 = (a^6 - b^3)(a^6 + b^3)\) - \(8x^3 - y^9 = (2x - y^3)((2x)^2 + 2xy^3 + (y^3)^2)\) - \(x^3 + 8y^3 = (x + 2y)(x^2 - 2xy + 4y^2)\)

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