In this chapter we will learn the concept & formula for cube of binomial. In the end we will also see the solved problems related to the concept.
Let us first understand the basics of binomial.
What is Binomial?
Binomial is an algebraic expression with two entities.
The entities can be variable, constant or both.
Examples of binomial are given below;
\mathtt{\Longrightarrow \ 6x\ +\ 5y}\\\ \\ \mathtt{\Longrightarrow \ 2\ +\ 3x}\\\ \\ \mathtt{\Longrightarrow \ 8x^{3} +9xy}\\\ \\ \mathtt{\Longrightarrow \ 12xy^{2} +x^{2} y}
Cube of Binomial formula
There are two important formula for cube of binomial;
(i) \mathtt{ \ \ ( a+b)^{3} =a^{3} +b^{3} +3ab\ ( a+b)}\\\ \\ \mathtt{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }
(ii) \mathtt{\ ( a-b)^{3} =a^{3} -b^{3} -3ab\ ( a-b)}\\\ \\ \mathtt{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }
You need to memorize both the formulas as they are important to solved cube of binomials problems.
Cube of Binomial solved problem
Example 01
Find the cube of binomial 10x – 3y
Solution
The cube of given binomial is expressed as \mathtt{( 10x-3y)^{3}}
Here we will use the below formula;
\mathtt{( a-b)^{3} =a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }
Here;
a = 10x
b = 3y
Putting the values;
\mathtt{\Longrightarrow \ ( 10x-3y)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 10x)^{3} -( 3y)^{3} -3( 10x)^{2}( 3y) +3( 10x)( 3y)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 1000x^{3} -27y^{3} -900x^{2} y+270xy^{2}}
The above expression is the cube of given binomial.
Example 02
Find the cube of binomial 5x + 2y
Solution
The cube of binomial is expressed as \mathtt{( 5x+2y)^{3}}
We will use the formula;
\mathtt{( a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }
Putting the values;
\mathtt{\Longrightarrow \ ( 5x+2y)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x)^{3} +( 2y)^{3} +3( 5x)^{2}( 2y) +3( 5x)( 2y)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 125x^{3} +8y^{3} +150x^{2} y+60xy^{2}}
The above expression is the cube of given binomial.
Example 03
Expand \mathtt{( x+2y)^{3} +( x-2y)^{3}}
Solution
The questions involve cube of two binomial.
We will solve the question in two steps.
First expand \mathtt{( x+2y)^{3}}
\mathtt{\Longrightarrow \ ( x+2y)^{3}}\\\ \\ \mathtt{\Longrightarrow ( x)^{3} +( 2y)^{3} +3( x)^{2}( 2y) +3( x)( 2y)^{2}}\\\ \\ \mathtt{\Longrightarrow x^{3} +8y^{3} +6x^{2} y+12xy^{2}}
Now expand \mathtt{( x-2y)^{3}}
\mathtt{\Longrightarrow \ ( x-2y)^{3}}\\\ \\ \mathtt{\Longrightarrow ( x)^{3} -( 2y)^{3} -3( x)^{2}( 2y) +3( x)( 2y)^{2}}\\\ \\ \mathtt{\Longrightarrow x^{3} -8y^{3} -6x^{2} y+12xy^{2}}
Now add both the parts;
\mathtt{\Longrightarrow \ ( x+2y)^{3} +( x-2y)^{3}}\\\ \\ \mathtt{\Longrightarrow x^{3} +\cancel{8y^{3}} +\cancel{6x^{2} y} +12xy^{2} +x^{3} -\cancel{8y^{3}} -\cancel{6x^{2} y} +12xy^{2}}\\\ \\ \mathtt{\Longrightarrow \ 12x^{3} +24xy^{2}}
Hence, the above expression is solution of given binomials.
Example 04
Find the value of \mathtt{( 19)^{3}} using cube of binomial formula.
Solution
The number can be converted into binomial as;
\mathtt{\Longrightarrow \ ( 19)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 20-1)^{3}}
We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }
Putting the values;
\mathtt{\Longrightarrow \ ( 20-1)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 20)^{3} -( 1)^{3} -3( 20)^{2}( 1) +3( 20)( 1)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 8000-1-1200\ +60}\\\ \\ \mathtt{\Longrightarrow \ 6859}
Hence, 6859 is the value of given number.
Example 05
Find the coefficient of \mathtt{x^{2} y} in the expression \mathtt{( 6x+y)^{3}}
Solution
We will use the formula;
\mathtt{( a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }
Putting the values;
\mathtt{\Longrightarrow \ ( 6x+y)^{3}}\\\ \\ \mathtt{\Longrightarrow ( 6x)^{3} +( y)^{3} +3( 6x)^{2}( y) +3( 6x)( y)^{2}}\\\ \\ \mathtt{\Longrightarrow 216x^{3} +y^{3} +78x^{2} y\ +18xy^{2}}
Here the coefficient of \mathtt{x^{2} y} is 78
Example 06
Expand \mathtt{\left(\frac{1}{2} x+y\right)^{3} -\left(\frac{1}{2} x-y\right)^{3}}
Solution
The question involve cube of two binomial.
We will solve the question in two steps;
First expand \mathtt{\left(\frac{1}{2} x+y\right)}
Using the formula;
\mathtt{( a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }
Put the values;
\mathtt{\Longrightarrow \ \left(\frac{1}{2} x+y\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \left(\frac{1}{2} x\right)^{3} +( y)^{3} +3\left(\frac{1}{2} x\right)^{2}( y) +3\left(\frac{1}{2} x\right)( y)^{2}}\\\ \\ \mathtt{\Longrightarrow \frac{\mathtt{x^{3}}}{8} +y^{3} +\frac{3}{4} x^{2} y\ +\frac{3}{2} xy^{2}}
Now expand \mathtt{\left(\frac{1}{2} x-y\right)^{3}}
We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }
Putting the values;
\mathtt{\Longrightarrow \ \left(\frac{1}{2} x-y\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \left(\frac{1}{2} x\right)^{3} -( y)^{3} -3\left(\frac{1}{2} x\right)^{2}( y) +3\left(\frac{1}{2} x\right)( y)^{2}}\\\ \\ \mathtt{\Longrightarrow \frac{\mathtt{x^{3}}}{8} -y^{3} -\frac{3}{4} x^{2} y\ +\frac{3}{2} xy^{2}}
Joining both the parts;
\mathtt{\Longrightarrow \ \left(\frac{1}{2} x+y\right)^{3} -\left(\frac{1}{2} x-y\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \cancel{\mathtt{\frac{\mathtt{x^{3}}}{8}}} +y^{3} +\frac{3}{4} x^{2} y\ +\cancel{\mathtt{\frac{3}{2} xy^{2}}} -\cancel{\mathtt{\frac{\mathtt{x^{3}}}{8}}} +y^{3} +\frac{3}{4} x^{2} y\ -\cancel{\mathtt{\frac{3}{2} xy^{2}}}}\\\ \\ \mathtt{\Longrightarrow \ 2y^{3} +\frac{6}{4} x^{2} y}
Hence, above expression is solution of given binomial