# Algebraic Manipulation Identities

An **identity** is an equality that holds true regardless of the values chosen for its variables.They are used in simplifying or rearranging algebra expressions. By definition, the two sides of an identity are interchangeable, so we can replace one with the other at any time. For example, the identity \((x+y)^2 = x^2 + 2xy + y^2\)
is true for all choices of \(x\) and \(y\), whether they are real or complex numbers.

#### Contents

## Technique

Since identities are true for all valid values of its variables, one side of the equality can be swapped for the other. For example, we can replace any instance of \((x+y)^2\) with \(x^2 + 2xy + y^2\) and vice versa because \((x+y)^2 = x^2 + 2xy + y^2\) is an identity.

Clever use of identities offers shortcuts to many problems by making the algebra easier to manipulate. Below are lists of some common algebraic identities.

These identities are product formulas that are basic examples of the binomial theorem:

\[\begin{align}

(x+y)^2 &= x^2 + 2xy + y^2 \\

(x-y)^2 &= x^2 - 2xy + y^2 \\

(x+y)^3 &= x^3+3x^2y + 3xy^2 + y^3 \\

(x-y)^3 &= x^3-3x^2y + 3xy^2 - y^3 \\

(x+y)^4 &= x^4 + 4x^3y + 6x^2y^2+4xy^3 + y^4 \\

(x-y)^4 &= x^4 - 4x^3y + 6x^2y^2-4xy^3 + y^4.
\end{align}\]

These identities are factoring formulas, and their more general forms are listed on the factorization page:

\[\begin{align}

x^2 - y^2 &= (x+y)(x-y) \\

x^3 - y^3 &= (x-y)\big(x^2+xy+y^2\big) \\

x^3 + y^3 &= (x+y)\big(x^2-xy+y^2\big) \\

x^4 - y^4 &= \big(x^2-y^2\big)\big(x^2+y^2\big).
\end{align}\]

## Application and Extensions

The identity \(4(x+7)(2x-1)=Ax^2+Bx+C\) holds for all real values of \(x\). What is \(A+B+C?\)

Multiplying out the left side of the identity we get

\[4(x+7)(2x-1)=8x^2+52x-28.\]

This expression must be equal to the right-hand side of the identity, so

\[8x^2+52x-28=Ax^2+Bx+C,\]

which gives \(A=8\), \(B=52\), and \(C=-28.\)

Therefore, \(A+B+C=8+52-28=32\). \(_\square\)

If \(A+B=8\) and \(AB=13\), what is \(A^3+B^3?\)

While you

couldsolve for \(A\) and \(B\), a more elegant solution exploits the identity\[(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3,\]

which can be rewritten as

\[x^3+y^3 = (x+y)^3-3xy(x+y).\]

Substituting in \(A\) and \(B\) for \(x\) and \(y,\) respectively, we get

\[\begin{align}

A^3+B^3 &= (A+B)^3-3AB(A+B) \\

&= (8)^3-3(13)(8) \\

&= 512-312 \\

&= 200.\ _\square

\end{align}\]

## Common Factorizations

**Factorization** is the decomposition of an expression into a product of its factors and is one of the central problems in algebra. The following are common factorizations:

For any positive integer \(n\), \[a^n-b^n = (a-b)\left(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1} \right).\] In particular, for \( n=2\), we have \( a^2-b^2=(a-b)(a+b)\).

For an odd positive integer \( n\), \[ a^n+b^n = (a+b)\left(a^{n-1} - a^{n-2} b + \cdots - ab^{n-2} + b^{n-1} \right).\]

\( a^2 \pm 2ab + b^2 = (a\pm b)^2.\)

\( x^3 + y^3 + z^3 - 3 xyz = (x+y+z) \left(x^2+y^2+z^2-xy-yz-zx\right).\)

\((ax+by)^2 + (ay-bx)^2 = \left(a^2+b^2\right)\left(x^2+y^2\right)\\ (ax+by)^2 - (ay+bx)^2 = \left(a^2-b^2\right)\left(x^2-y^2\right).\)

\( x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x).\)

Factorization often transforms an expression into a form that is more easily manipulated algebraically, has easily recognizable solutions, and gives rise to clearly defined relationships.

## Algebraic Manipulation Identities - Basic

## Find all ordered pairs of positive integer solutions \( (x,y)\) such that \(2^x+ 1 = y^2\).

We have \(2^x = y^2-1 = (y-1)(y+1)\). Since the factors \((y-1)\) and \((y+1)\) on the right-hand side are integers whose product is a power of 2, both \((y-1)\) and \((y+1)\) must be powers of 2. Furthermore, their difference is

\[ (y+1)-(y-1)=2,\]

implying the factors must be \(y+1 = 4\) and \(y-1 = 2\). This gives \( y=3\), and thus \(x=3\). Therefore, \((3, 3)\) is the only solution. \(_\square\)

## Algebraic Manipulation Identities - Intermediate

One way to discover factors is to find values for which the expression is equal to zero and apply the remainder-factor theorem. For example, in expression (1) above, \(a^n - b^n = 0\) for \(a=b\), implying \(a-b\) is a factor of \(a^n - b^n\). Similarly, in expression (2) for \(n\) odd, \(a^n - b^n = 0\) for \(a=-b\), implying that \(a+b\) is a factor of \(a^n+b^n\). An expression that can be factored into products of smaller degree is called *reducible*; otherwise, the expression is *irreducible*.

Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.

For any integer \(n\), prove that \(f(n)= n^5-5n^3 + 4n \) is divisible by \(120\).

We can factorize to obtain

\[\begin{align} f(n) & = n(n^4 - 5n^2 + 4) \\ &= n(n^2 -4)(n^2-1) \\ &= n(n-2)(n+2)(n-1)(n+1) \\ &= (n-2)(n-1)n(n+1)(n+2). \end{align} \]

Now, for any integer \(n\), the right-hand side is the product of five consecutive integers. One of these integers is divisible by 5, at least one of these integers is divisible by 3, and at least two of these integers are divisible by 2, one of which is divisible by 4. Therefore, \(f(n)\) is divisible by \(5 \cdot 4 \cdot 3 \cdot 2 = 120\). \(_\square\)

Factorize the polynomial \( \left( 1 + x + x^2 + \cdots + x^n \right)^2 - x^n\).

Let \(P(x) = 1 + x + x^2 + \cdots + x^{n-1}\). Then we have

\[\begin{align} \big( 1 + x + x^2 + \cdots + x^n \big)^2 - x^n &= (P(x) + x^n)^2 - x^n \\ &= P(x)^2 + 2P(x) x^n + x^{2n} - x^n\\ &= P(x)^2 + 2P(x) x^n + (x^{n} - 1)x^n\\ &= P(x)^2 + 2P(x) x^n + P(x) (x-1)x^n\\ &= P(x) (P(x) + x^n + x^{n+1})\\ &= \big(1 + x + x^2 + \cdots + x^{n-1}\big) \big(1 + x + x^2 + \cdots + x^n + x^{n+1}\big). \ _\square \end{align}\]

Factorize the polynomial \[f(a, b, c) = ab\big(a^2-b^2\big) + bc\big(b^2-c^2\big) + ca\big(c^2-a^2\big).\]

Observe that if \( a=b\), then \(f(a, a, c) =0\); if \(b=c\), then \(f(a, b, b)=0\); and if \( c=a\), then \( f(c,b,c)=0\). By the remainder-factor theorem, \( (a-b), (b-c),\) and \( (c-a)\) are factors of \( f(a,b,c)\). This allows us to factorize

\[f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).\ _\square\]

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