Question

Find the cube of the term \[\left( 2x+5 \right)\]?

Answer 1

Hint: This question is from the topic of algebra. In solving this question, we will first find out the square of the term \[\left( 2x+5 \right)\]. We will find the square using the foil method. After that, we will multiply the squared term with the term \[\left( 2x+5 \right)\] to get the cube of \[\left( 2x+5 \right)\]. Here also, we will use the foil method to solve the question further and get the answer. After that, we will see an alternate method to solve this question.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find the cube of \[\left( 2x+5 \right)\]. Or, we can say we have to find the value of \[{{\left( 2x+5 \right)}^{3}}\].
As we know that we can write \[{{x}^{3}}\] as \[x\cdot x\cdot x\] or \[{{x}^{2}}\cdot x\], so we can write
\[{{\left( 2x+5 \right)}^{3}}={{\left( 2x+5 \right)}^{2}}\cdot \left( 2x+5 \right)\]
Let us first solve the term \[{{\left( 2x+5 \right)}^{2}}\].
The term \[{{\left( 2x+5 \right)}^{2}}\] can also be written as
\[{{\left( 2x+5 \right)}^{2}}=\left( 2x+5 \right)\cdot \left( 2x+5 \right)\]
Now, using foil method formula that is (a+b)(c+d)=ac+ad+bc+bd, we can write
\[\Rightarrow {{\left( 2x+5 \right)}^{2}}=2x\cdot 2x+2x\cdot 5+5\cdot 2x+5\cdot 5\]
The above equation can also be written as
\[\Rightarrow {{\left( 2x+5 \right)}^{2}}={{\left( 2x \right)}^{2}}+10x+10x+{{5}^{2}}\]
The above equation can also be written as
\[\Rightarrow {{\left( 2x+5 \right)}^{2}}=4{{x}^{2}}+20x+25\]
Now, let us find the value of \[{{\left( 2x+5 \right)}^{3}}\]. This can also be written as
\[{{\left( 2x+5 \right)}^{3}}={{\left( 2x+5 \right)}^{2}}\cdot \left( 2x+5 \right)\]
The above equation can also be written as
\[\Rightarrow {{\left( 2x+5 \right)}^{3}}=\left( 4{{x}^{2}}+20x+25 \right)\cdot \left( 2x+5 \right)\]
The above equation can also be written as
\[\Rightarrow {{\left( 2x+5 \right)}^{3}}=4{{x}^{2}}\cdot 2x+4{{x}^{2}}\cdot 5+20x\cdot 2x+20x\cdot 5+25\cdot 2x+25\cdot 5\]
The above equation can also be written as
\[\Rightarrow {{\left( 2x+5 \right)}^{3}}=8{{x}^{3}}+20{{x}^{2}}+40{{x}^{2}}+100x+50x+125\]
\[\Rightarrow {{\left( 2x+5 \right)}^{3}}=8{{x}^{3}}+60{{x}^{2}}+150x+125\]
So, we have found the cube of \[\left( 2x+5 \right)\]. The cube of \[\left( 2x+5 \right)\] is \[8{{x}^{3}}+60{{x}^{2}}+150x+125\].

Note: We should have a better knowledge in the topic of algebra. We should know that the \[{{x}^{a+b}}\] can be written as \[{{x}^{a}}{{x}^{b}}\].
We can solve this question by alternate method.
For solving this question by alternate method, we should know the formula of \[{{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\]
Now, using this formula, we can write
\[{{\left( 2x+5 \right)}^{3}}={{\left( 2x \right)}^{3}}+3\cdot {{\left( 2x \right)}^{2}}\cdot 5+3\cdot (2x)\cdot {{5}^{2}}+{{5}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( 2x+5 \right)}^{3}}=\left( 8{{x}^{3}} \right)+3\cdot \left( 4{{x}^{2}} \right)\cdot 5+3\cdot (2x)\cdot 25+125\]
\[\Rightarrow {{\left( 2x+5 \right)}^{3}}=8{{x}^{3}}+60{{x}^{2}}+150x+125\]
As we have got the same answer, so we can use this method too to solve this question.