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Mei Li, Mahindra Jain, Niranjan Khanderia, and 5 others

• Andrew Hayes
• Arron Kau
• Satyabrata Dash
• Andres Gonzalez
• Jimin Khim

contributed

An algebraic identity is an equality that holds for any values of its variables.

For example, the identity \[(x+y)^2 = x^2 + 2xy + y^2\] holds for all values of \(x\) and \(y\).

Since an identity holds for all values of its variables, it is possible to substitute instances of one side of the equality with the other side of the equality. For example, because of the identity above, we can replace any instance of \((x+y)^2\) with \(x^2 + 2xy + y^2\) and vice versa.

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

Contents

• Binomial Theorem Identities
• Factoring Identities
• Three-variable Identities
• Example Problems

The following identities are product formulas that are examples of the binomial theorem:

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

The following identities are factoring formulas; more general factoring formulas are listed in the wiki Algebraic Manipulation - Identities:

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

The following identities can be derived by some clever factoring and manipulation of the terms:

\[\begin{align} (x+y)(x+z)(y+z) &= (x+y+z)(xy+xz+yz)-xyz \\ x^2+y^2+z^2 &= (x+y+z)^2-2(xy+xz+yz) \\ x^3+y^3+z^3-3xyz &= (x+y+z)(x^2+y^2+z^2-xy-xz-yz). \end{align}\]

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 have \[4(x+7)(2x-1)=8x^2+52x-28.\] This expression must be equal to the right hand side of the identity, implying \[8x^2+52x-28=Ax^2+Bx+C,\] so \(A=8\), \(B=52\), and \(C=-28\), and \(A+B+C=8+52-28=32\). \(_\square\)

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

While it is possible to solve 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 \(y,\) we get \[\begin{align} A^3+B^3 &= (A+B)^3-3AB(A+B) \\ &= (8)^3-3(13)(8) \\ &= 512-312 \\ &= 200.\ _\square \end{align}\]

The following identity holds for all real numbers \(x\) and \(y:\) \[-7x+4y+11=a(x-2y+1)+b(-2x+3y+3)+c.\] What is \(a+b+c?\)

Rewriting the identity, we have \[\begin{align} -7x+4y+11&=a(x-2y+1)+b(-2x+3y+3)+c\\ -7x+4y+11&=(a-2b)x+(-2a+3b)y+(a+3b+c). \end{align}\] This gives the following system of equations: \[\begin{align} a-2b&=-7\\ -2a+3b&=4\\ a+3b+c&=11. \end{align}\] Solving this system of equations gives \[a=13, b=10, c=-32 \implies a+b+c=-9. \ _\square\]

For all real numbers \(x\) and \(y\) such that \(x-y=3,\) the following identity holds: \[axy+bx+cy+9=0.\] What is \(a+b+c?\)

Since \(x-y=3\) implies \(y=x-3,\) substituting this into the given identity gives \[\begin{align} ax(x-3)+bx+c(x-3)+9&=0\\ ax^2+(-3a+b+c)x-3(c-3)&=0. \end{align}\] Since this is an identity in \(x,\) we have \[a=0, c=3, b=-3 \implies a+b+c=0. \ _\square\]

If \(x+y=12\) and \(xy=35,\) what is \(x^4+y^4?\)

Since \(x^4+y^4=\big(x^2+y^2\big)^2-2x^2y^2\) holds for all real values of \(x\) and \(y,\) we have \[\begin{align} x^4+y^4 &=\big(x^2+y^2\big)^2-2x^2y^2\\ &=\big((x+y)^2-2xy\big)^2-2(xy)^2\\ &=(12^2-2\cdot35)^2-2\cdot35^2\\ &=5476-2450\\ &=3026. \ _\square \end{align}\]

If \(a+b+c=6, a^2+b^2+c^2=14,\) and \(ab+bc+ca=11,\) what is \(a^3+b^3+c^3-3abc?\)

Substituting \(a+b+c=6, a^2+b^2+c^2=14,\) and \(ab+bc+ca=11\) into the identity \[a^3+b^3+c^3-3abc=(a+b+c)\big(a^2+b^2+c^2-ab-bc-ca\big),\] we have \(a^3+b^3+c^3-3abc=6\cdot(14-11)=18.\) \(_\square\)

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