Question

What are the cubic formulae used in factorization?

Answer 1

Answer:

a^3+b^3=(a+b)(a^2-ab+b^2)

Step-by-step explanation:

Answer 2

For a general Cubic Equation:

$a {x}^{3} + b {x}^{2} + cx + d = 0$

${x}^{3} + \frac{b}{a} {x}^{2} + \frac{c}{a} x + \frac{d}{a} = 0$

${(x+\frac{b}{3a})}^{3} + p(x+\frac{b}{3a}) + q = 0$

Where,

$p = \frac{3ac - {b}^{2} }{3 {a}^{2} } \\ \: q = \frac{2 {b}^{3} - 9abc + 27 {a}^{2}d }{27 {a}^{3} }$

All the Three Roots are:

$\\ x _{1} = -\frac{b}{3a}+ \sqrt[3]{ - \frac{ q}{2} + \sqrt{ {( \frac{q}{2} )}^{2} + ( \frac{p}{3}) ^{3} } } + \sqrt[3]{ - \frac{q}{2} - \sqrt{ {( \frac{q}{2} )}^{2} + {( \frac{p}{3}) }^{3} } }$

$\\ x _{2} = -\frac{b}{3a} + \omega · \sqrt[3]{ - \frac{ q}{2} + \sqrt{ {( \frac{q}{2} )}^{2} + ( \frac{p}{3}) ^{3} } } + \omega ^{2} · \sqrt[3]{ - \frac{q}{2} - \sqrt{ {( \frac{q}{2} )}^{2} + {( \frac{p}{3}) }^{3} } }$

$\\ x _{3} = -\frac{b}{3a} + \omega^{2} · \sqrt[3]{ - \frac{ q}{2} + \sqrt{ {( \frac{q}{2} )}^{2} + ( \frac{p}{3}) ^{3} } } + \omega · \sqrt[3]{ - \frac{q}{2} - \sqrt{ {( \frac{q}{2} )}^{2} + {( \frac{p}{3}) }^{3} } }$

Where, ω is one of the two non-real complex Cube roots of One.

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