Final Answer
$\frac{147}{4}+16\left(a-\frac{7}{8}\right)^2$
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Step-by-step Solution
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1
Factor the sum or difference of cubes using the formula: $a^3\pm b^3 = (a\pm b)(a^2\mp ab+b^2)$
$\frac{\left(4a+7\right)\left(16a^{2}-28a+49\right)}{4a+7}$
Learn how to solve polynomial factorization problems step by step online.
$\frac{\left(4a+7\right)\left(16a^{2}-28a+49\right)}{4a+7}$
Learn how to solve polynomial factorization problems step by step online. Factor by completing the square (64a^3+343)/(4a+7). Factor the sum or difference of cubes using the formula: a^3\pm b^3 = (a\pm b)(a^2\mp ab+b^2). Simplify the fraction \frac{\left(4a+7\right)\left(16a^{2}-28a+49\right)}{4a+7} by 4a+7. Use the complete the square method to factor the trinomial of the form ax^2+bx+c. Take common factor a (16) to all terms. Add and subtract \displaystyle\left(\frac{b}{2a}\right)^2.
Final Answer
$\frac{147}{4}+16\left(a-\frac{7}{8}\right)^2$