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# Proving Polynomial Identities

Equations that are true for every possible value of the variable are called identities. Some special polynomial equations are identities. Using polynomial identities can be useful when rewriting polynomial expressions. To prove an equation is an identity, it is enough to show that both sides can be written in the same way.
Rule

## Square of a Binomial

When a binomial is squared, the resulting expression is a perfect square trinomial.

### Rule

$(a + b)^2=a^2 + 2ab + b^2$
This identity can be shown by first rewriting the square as a product.
$(a + b)^2$
$(a + b)(a + b)$
$a\cdot a + a\cdot b + b\cdot a +b\cdot b$
$a^2 + ab + ab +b^2$
$a^2 + 2ab +b^2$

### Rule

$(a - b)^2=a^2 - 2ab + b^2$
In this case, when one term of the binomial is subtracted from the other, the middle term of the perfect square trinomial will instead be negative.
$(a - b)^2$
$(a - b)(a - b)$
$a\cdot a - a\cdot b - b\cdot a +b\cdot b$
$a^2 - ab - ab +b^2$
$a^2 - 2ab +b^2$
fullscreen
Exercise

Rewrite $(x+3)^2$ and $(7-x)^2$ as perfect square trinomials.

Show Solution
Solution
When rewriting $(x+3)^2$ as a perfect square trinomial, keep in mind that it is the square of a binomial.
$(x+3)^2$
$x^2+2\cdot x\cdot3+3^2$
$x^2+6x+3^2$
$x^2+6x+9$
The second expression has a minus sign between the terms – take this into account when rewriting the binomial.
$(7-x)^2$
$7^2-2\cdot7\cdot x +x^2$
$49-2\cdot7\cdot x +x^2$
$49-14x +x^2$
The expressions rewritten as perfect square trinomials are $x^2+6x+9$ and $49-14x +x^2,$ respectively.
Rule

## Product of a Conjugate Pair of Binomials

When a binomial is multiplied by its conjugate, the resulting expression is the difference of two squares.

### Rule

$(a + b)(a - b)=a^2 - b^2$
This identity can be shown using the Distributive Property — multiplying each term in the first set of parentheses with each term in the second.
$(a + b)(a - b)$
$a(a+b)-b(a+b)$
$a^2+ab-b(a+b)$
$a^2+ab-ba-b^2$
Simplify
$a^2+ab-ab-b^2$
$a^2 - b^2$
Thus, the product of a binomial and its conjugate is the difference of two squares.
Rule

## Sum of Two Cubes

### Rule

$a^3 + b^3 =(a+b)\left( a^2 - ab + b^2 \right)$
To show the identity $a^3 + b^3 =(a+b)\left( a^2 - ab + b^2 \right),$ it is enough to rewrite the right-hand side and show that it equals the sum on the left-hand side.
$(a+b)\left( a^2 - ab + b^2 \right)$
$a\cdot a^2 -a\cdot ab + a\cdot b^2 + b\cdot a^2 - b\cdot ab +b\cdot b^2$
$a^3 -a^2b + ab^2 + a^2b - ab^2 + b^3$
$a^3 - a^2b + a^2b + ab^2 - ab^2 + b^3$
$a^3 + b^3$
Rule

## Difference of Two Cubes

### Rule

$a^3 - b^3 =(a-b)\left( a^2 + ab + b^2 \right)$
Showing this identity is easiest done by rewriting the right-hand side.
$(a-b)\left( a^2 + ab + b^2 \right)$
$a\cdot a^2 +a\cdot ab + a\cdot b^2 - b\cdot a^2 - b\cdot ab -b\cdot b^2$
$a^3 +a^2b + ab^2 - a^2b - ab^2 - b^3$
$a^3 +a^2b - a^2b + ab^2 - ab^2 - b^3$
$a^3 - b^3$
Thus, the identity $a^3 - b^3 =(a-b)\left( a^2 + ab + b^2 \right)$ is true.
fullscreen
Exercise

Find all complex solutions to the equation $x^3-4^3=0$ by using the difference of two cubes.

Show Solution
Solution
To solve the equation $x^3-4^3=0,$ notice that the expression on the left-hand side is the difference of two cubes. We can use that to rewrite it as a product. $\begin{gathered} x^3-4^3=0 \\ \Updownarrow \\ (x-4) \left( x^2+x\cdot 4+4^2 \right) =0 \end{gathered}$ The Zero Product Property tells us that we can solve the equation by setting each of the factors equal to $0$ and solving the resulting equations separately. $x-4=0 \quad \text{and} \quad x^2+x\cdot 4+4^2 =0$ Let's find the solution to the first equation. $x-4=0 \quad \Leftrightarrow \quad x=4$ The second equation we will solve using the Quadratic Formula.
$x^2+x\cdot 4+4^2 =0$
$x^2+4x+4^2 =0$
$x^2+4x+16 =0$
$x=\dfrac{\text{-} {\color{#009600}{4}}\pm\sqrt{{\color{#009600}{4}}^2-4\cdot {\color{#0000FF}{1}}\cdot {\color{#FF0000}{16}}}}{2\cdot {\color{#0000FF}{1}}}$
$x=\dfrac{\text{-} 4\pm\sqrt{16-64}}{2}$
$x=\dfrac{\text{-} 4\pm\sqrt{\text{-}48}}{2}$
$x=\dfrac{\text{-} 4\pm i\sqrt{48}}{2}$
$\begin{array}{l}x=\frac{\text{-} 4+i\sqrt{48}}{2} \\ x=\frac{\text{-} 4-i\sqrt{48}}{2} \end{array}$
We should simplify these expressions, if possible. The real part becomes $\frac{\text{-} 4}{2}=\text{-} 2$ and the imaginary part can be simplified.
$\dfrac{i\sqrt{48}}{2}$
$\dfrac{i\sqrt{12\cdot 4}}{2}$
$\dfrac{i\sqrt{12}\cdot \sqrt{4}}{2}$
$\dfrac{i\sqrt{12}\cdot 2}{2}$
$i\sqrt{12}$
After we have simplified, we the complex solutions to the quadratic equation can be written as $x=\text{-} 2 + i\sqrt{12}$ and $x=\text{-} 2 - i\sqrt{12}.$ $\small x=4, \, \, x=\text{-} 2 + i\sqrt{12} \, \, \text{and} \, \, x=\text{-} 2 - i\sqrt{12} \small$
Rule

## Square of a Sum of Two Squares

### Rule

$\left(a^2 + b^2 \right)^2=\left( a^2 - b^2 \right)^2 + (2ab)^2$
To prove the identity $\left( a^2+b^2 \right)^2=\left( a^2 - b^2 \right)^2 + \left( 2ab \right)^2,$ it is enough to show that the right-hand side can be rewritten to equal the expression on the left-hand side.
$\left( a^2 - b^2 \right)^2 + \left( 2ab \right)^2$
$\left( a^2 \right)^2 - 2a^2b^2 + \left( b^2 \right)^2 + \left( 2ab \right)^2$
$\left( a^2 \right)^2 - 2a^2b^2 + \left( b^2 \right)^2 + 2^2a^2b^2$
$\left( a^2 \right)^2 - 2a^2b^2 + \left( b^2 \right)^2 + 4a^2b^2$
$\left( a^2 \right)^2 + 4a^2b^2 - 2a^2b^2 + \left( b^2 \right)^2$
$\left( a^2 \right)^2 + 2a^2b^2 + \left( b^2 \right)^2$
The three terms can be factored, as they form a perfect square trinomial.
$\left( a^2 \right)^2 + 2a^2b^2 + \left( b^2 \right)^2$
$\left( a^2+b^2 \right)^2$
By that, the proof of the identity is complete. $\left(a^2 + b^2 \right)^2=\left( a^2 - b^2 \right)^2 + (2ab)^2$