Page:Elementary algebra (1896).djvu/100

82 101. Important Cases. By a slight modification some ex- pressions admit of being written in the form of the difference of two squares, and may then be resolved into factors by the method of Art. 98.

Ex. 1. Resolve into factors $$x^4 + {x^2}{y^2} + y^4$$

$$ \begin{align} x^4 + {x^2}{y^2} + y^4 &= x \left ( x^4 + 2{x^2}{y^2} \right ) {x^2}{y^2} \\ &= \left ( {x^2} + {y^2}^2 \right ) - \left ( {xy}^2 \right ) \\ &= \left ( {x^2} + {y^2} + xy \right ) \left ( x^2 + y2 - xy \right ) \\ &= \left ( {x^2} + {y^2} + xy \right ) \left ( x^2 + y2 - xy \right ) \\ &= \left ( {x^2} + xy + {y^2} \right ) \left ( x^2 - xy + y2 \right )  \\ \end{align} $$

Ex. 2. Resolve into factors $$x^4 - 15{x^2}{y^2} + 9y^4$$.

$$ \begin{align} {x^4} - 15 {x^2}{y^2} + 9{y^4} &= \left ( {x^4} - 6{x^2}{y^2} + 9{y^4} \right ) - 9{x^2}{y^2} \\ &= { \left ( x^2-3y^2 \right ) }^2 - { \left (3xy \right ) }^2 \\ &= \left ( x^2 - 3y^2 + 3xy \right ) \left ( x^2 - 3y^2 - 3xy \right ) \end{align} $$

Resolve into factors :


 * 1) $$x^4 + 16x^2 + 256.$$.
 * 2) $$81a^4 + 9a^2b^2 + b^4.$$.
 * 3) $$x^4 + y^4 - 7 x^2y2$$.
 * 4) $$m^4 + n^4 - 18 {m^2}{n^2}$$.
 * 5) $$x^4 - 6{x^2}{y^2} + y^4$$.
 * 6) $$4{x^4} + 9{y^4} - 93{x^2}{y^2}$$.
 * 7) $$4{m^4} + 9{n^4} - 24{m^2}{n^2}$$.
 * 8) $$9{x^4} + 4{y^4} + 11{x^2}{y^2}$$.
 * 9) $$x^4 - 19{x^2}{y^2} + 25 {y^4}$$.
 * 10) $$16a^4 + b^4 - 28{a^2 b^2}$$.

102. If we divide $$a^3 + b^3$$ by $$a + b$$ the quotient is $$a^2 = ab + b^2 $$; and if we divide $$a^3 - b^3$$ by $$a - b$$ the quotient is $$a^2 + ab +b^2$$. We have therefore the following identities:

$$\begin{align} a^3 + b^3 &= \left (a + b \right ) \left (a^2 - ab + b^2 \right ) ; \\ a^3 - b^3 &= \left (a - b \right ) \left (a^2 + ab + b^2 \right ). \\ \end{align} $$

These results are very important, and enable us to resolve into factors any expression which can be written as the sum or the difference of two cubes.

Ex. 1. $$\begin{align} 8x^3 - 27y^3 &= { \left ( 2x \right ) }^3 - { \left ( 3y \right ) }^3 \\ &= { \left (  4x^2 + 6xy +9y^2 \right ) } \\ \end{align} $$

The middle term $$6 xy$$ is the product of $$2x$$ and $$3y$$.