
How do you factor \[{\left( {a + 3b} \right)^3} - {\left( {2a + 3b} \right)^3}\] ?
Answer
489.9k+ views
Hint: Given are the expressions with two terms in each bracket. We will use the standard cubic expansion identity of the bracket above. For two separate terms. Then we will take the terms together that have the same coefficient. This will simplify the terms. also if any term is common then take that outside.
Complete step by step answer:
Given expression is,\[{\left( {a + 3b} \right)^3} - {\left( {2a + 3b} \right)^3}\].Now we will use the standard cubic expansion identity,
\[{\left( {a + b} \right)^3} = {a^3} + 3a{b^2} + 3{a^2}b + {b^3}\]
\[\Rightarrow {a^3} + 3a{\left( {3b} \right)^2} + 3{a^2} \times 3b + {\left( {3b} \right)^3} - \left( {8{a^3} + 3 \times 2a \times {{\left( {3b} \right)}^2} + 3{{\left( {2a} \right)}^2} \times 3b + 27{b^3}} \right)\]
Taking the respective cubes and squares,
\[{a^3} + 3a \times 9{b^2} + 9{a^2}b + 27{b^3} - \left( {8{a^3} + 6a \times 9{b^2} + 9b \times 4{a^2} + 27{b^3}} \right)\]
Multiply the respective constants,
\[{a^3} + 27a{b^2} + 9{a^2}b + 27{b^3} - \left( {8{a^3} + 54a{b^2} + 36b{a^2} + 27{b^3}} \right)\]
Multiplying the terms inside the bracket with minus sign,
\[{a^3} + 27a{b^2} + 9{a^2}b + 27{b^3} - 8{a^3} - 54a{b^2} - 36b{a^2} - 27{b^3}\]
Now take the terms with same coefficient together,
\[{a^3} - 8{a^3} + 27a{b^2} - 54a{b^2} + 9{a^2}b - 36{a^2}b\]
\[\Rightarrow - 7{a^3} - 27a{b^2} - 27{a^2}b\]
Taking -a common,
\[\therefore - a\left( {7{a^2} + 27{b^2} + 27ab} \right)\]
These are the factors of the above expression.
Therefore, the factor of \[{\left( {a + 3b} \right)^3} - {\left( {2a + 3b} \right)^3}\] is $- a\left( {7{a^2} + 27{b^2} + 27ab} \right)$.
Note: Note that terms with the same coefficient only can be added or subtracted together. This is not restricted for multiplication and division. Also note that when we multiply the second bracket with minus sign outside it the signs will change. This question can be solved by one more method.We can use standard expansion identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} - ab + {b^2}} \right)\]. Here $a$ can be replaced by the first bracket and $b$ will be replaced by the second bracket.
Complete step by step answer:
Given expression is,\[{\left( {a + 3b} \right)^3} - {\left( {2a + 3b} \right)^3}\].Now we will use the standard cubic expansion identity,
\[{\left( {a + b} \right)^3} = {a^3} + 3a{b^2} + 3{a^2}b + {b^3}\]
\[\Rightarrow {a^3} + 3a{\left( {3b} \right)^2} + 3{a^2} \times 3b + {\left( {3b} \right)^3} - \left( {8{a^3} + 3 \times 2a \times {{\left( {3b} \right)}^2} + 3{{\left( {2a} \right)}^2} \times 3b + 27{b^3}} \right)\]
Taking the respective cubes and squares,
\[{a^3} + 3a \times 9{b^2} + 9{a^2}b + 27{b^3} - \left( {8{a^3} + 6a \times 9{b^2} + 9b \times 4{a^2} + 27{b^3}} \right)\]
Multiply the respective constants,
\[{a^3} + 27a{b^2} + 9{a^2}b + 27{b^3} - \left( {8{a^3} + 54a{b^2} + 36b{a^2} + 27{b^3}} \right)\]
Multiplying the terms inside the bracket with minus sign,
\[{a^3} + 27a{b^2} + 9{a^2}b + 27{b^3} - 8{a^3} - 54a{b^2} - 36b{a^2} - 27{b^3}\]
Now take the terms with same coefficient together,
\[{a^3} - 8{a^3} + 27a{b^2} - 54a{b^2} + 9{a^2}b - 36{a^2}b\]
\[\Rightarrow - 7{a^3} - 27a{b^2} - 27{a^2}b\]
Taking -a common,
\[\therefore - a\left( {7{a^2} + 27{b^2} + 27ab} \right)\]
These are the factors of the above expression.
Therefore, the factor of \[{\left( {a + 3b} \right)^3} - {\left( {2a + 3b} \right)^3}\] is $- a\left( {7{a^2} + 27{b^2} + 27ab} \right)$.
Note: Note that terms with the same coefficient only can be added or subtracted together. This is not restricted for multiplication and division. Also note that when we multiply the second bracket with minus sign outside it the signs will change. This question can be solved by one more method.We can use standard expansion identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} - ab + {b^2}} \right)\]. Here $a$ can be replaced by the first bracket and $b$ will be replaced by the second bracket.
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