# Formula for a^3 + b^3 + c^3 – 3abc

In this chapter, we will discuss one important algebra formula that is relevant for grade 9 math student.

The formula is given as;

\mathtt{( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}\\\ \\ \mathtt{\Longrightarrow \ a^{3} +b^{3} +c^{3} -3abc}

This formula is important and you need to memorize it as it is directly asked in the exams.

Given below is the proof of the formula;

\mathtt{( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}

Multiplying individual terms;

\mathtt{\Longrightarrow \ \left( a^{3} +ab^{2} +ac^{2} -a^{2} b-abc-ca^{2}\right)}\\ \\ \mathtt{+\ \left( a^{2} b+b^{3} +bc^{2} -ab^{2} -b^{2} c-abc\right)}\\ \\ \mathtt{+\ \left( a^{2} c+b^{2} c+c^{3} -abc -bc^{2} -ac^{2}\right)}

Cancelling common terms.

\mathtt{\Longrightarrow \ \left( a^{3} +\cancel{ab^{2}} +\cancel{ac^{2}} -\cancel{a^{2} b} -abc-\cancel{ca^{2}}\right)}\\ \\ \mathtt{+\ \left(\cancel{a^{2} b} +b^{3} +\cancel{bc^{2}} -\cancel{ab^{2}} -\cancel{b^{2} c} -abc\right)}\\ \\ \mathtt{+\ \left(\cancel{ca^{2}} +\cancel{b^{2} c} +c^{3} -abc -\cancel{bc^{2}} -\cancel{ac^{2}}\right)}\\\ \\ \mathtt{\Longrightarrow \ a^{3} +b^{3} +c^{3} -3abc}

I hope you understood the above formula. Given below are some solved problems for further clarity.

Example 01
Solve the below expression.

\mathtt{( 5x+2y+3z) \ \left( 25x^{2} +4y^{2} +9z^{2} -10xy-6yz-15xz\right)}

Solution
Referring the formula;

\mathtt{( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}\\ \\ \mathtt{=a^{3} +b^{3} +c^{3} -3abc}

Where;
a = 5x
b = 2y
c = 3z

Using the formula we get;

\mathtt{\Longrightarrow \ ( 5x)^{3} +( 2y)^{3} +( 3z)^{3} -3( 5x)( 2y)( 3z)}\\\ \\ \mathtt{\Longrightarrow \ 125x^{3} +8y^{3} +27z^{3} -90xyz}

Hence, the above expression is the solution.

Example 02
Simplify the expression;

\mathtt{( x-y+6z) \ \left( x^{2} +y^{2} +36z^{2} +xy+6yz-6xz\right)}

Solution
Referring the formula;

\mathtt{( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}\\ \\ \mathtt{=a^{3} +b^{3} +c^{3} -3abc}

Where;
a = x
b = -y
c = 6z

Using the formula, we get;

\mathtt{\Longrightarrow \ ( x)^{3} +( -y)^{3} +( 6z)^{3} -3( x)( -y)( 6z)}\\\ \\ \mathtt{\Longrightarrow \ x^{3} -y^{3} +216z^{3} +18xyz}

Hence, the above expression is the solution.

Example 03
Simplify the expression;
\mathtt{( x-3-4y) \ \left( x^{2} +9+16y^{2} +3x-12y+4xy\right)}

Solution
Referring the formula;

\mathtt{( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}\\ \\ \mathtt{=a^{3} +b^{3} +c^{3} -3abc}

Where;
a = x
b = -3
c = -4y

Using the formula, we get;

\mathtt{\Longrightarrow \ ( x)^{3} +( -3)^{3} +( -4y)^{3} -3( x)( -3)( -4y)}\\\ \\ \mathtt{\Longrightarrow \ x^{3} -9-64y^{3} -36xy}

Hence, the above expression is the solution.

Example 04
Expand the below expression;

\mathtt{\Longrightarrow \ 1000+8x^{3} +y^{3} -60xy}

Solution
The number can be expressed as;

\mathtt{\Longrightarrow \ ( 10)^{3} +( 2x)^{3} +y^{3} -3( 10)( 2x)( y)}

Referring the formula;

\mathtt{a^{3} +b^{3} +c^{3} -3abc}\\\ \\ \mathtt{=\ ( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}

Where;
a = 10
b = 2x
c = y

Now using the above formula, expanding the expression;

\mathtt{\Longrightarrow \ ( 10+2x+y)\left( 10^{2} +( 2x)^{2} +y^{2} -10.2x-2xy-10y\right)}\\\ \\ \mathtt{\Longrightarrow \ ( 10+2x+y)\left( 100+4x^{2} +y^{2} -20x-2xy-10y\right)}

Hence, the above expression is the solution.

Example 05
Solve the below expression.

\mathtt{\ ( -7x+5y-9z)\left( 49x^{2} +25y^{2} +81z^{2} +35xy+45yz-63zx\right)}

Solution
Referring the formula;

\mathtt{( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ca\right)}\\ \\ \mathtt{=a^{3} +b^{3} +c^{3} -3abc}

Where;
a = -7x
b = 5y
c = -9z

Using the formula, we get;

\mathtt{\Longrightarrow \ ( -7x)^{3} +( 5y)^{3} +( -9z)^{3} -( -7x)( 5y) -( 5y)( -9z) -( -9z)( -7x)}\\\ \\ \mathtt{\Longrightarrow \ -343x^{3} +125y^{3} -729z^{3} +35xy+45yz-63zx}

Hence, the above expression is the solution.

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