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solve by factoring x^3 - 64=0
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x^3 - 64=0 x^3 = 64 \[x \sqrt[3]{64} = 4\]
\[x = \sqrt[3]{64} = 4\]
has to be by factoring you have to use SOAP then use quadratic formula
difference of cubes... has 3 roots, 1 real and 2 complex
\(\tt \color{black}{\text{use formula}}\) \(\large\tt \color{black}{a^3-b^3=(a-b)(a^2+ab+b^2)}\)
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i used the formula. now i have to find the Zeros (answers)
\(\tt \color{black}{\text{u will get one solution x=4 thats clear}}\) \(\large \tt \color{blue}{for ~~x^2+4x+16=0}\) \(\tt \color{black}{\text{use formula}}\) \(\large \tt \color{blue}{x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}\)
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