How do you simplify ${\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$?
Answer
609.9k+ views
Hint: We can observe that $a$ is common in both the terms in the expression so we can take that out to make the expression slightly simpler. Then we will apply the algebraic formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. On further simplification, this will give us the answer.
Complete step by step answer:
According to the question, we have to show the process of simplification of an expression ${\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$.
Let this expression be denoted by a variable $k$. So we have:
$ \Rightarrow k = {\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$
We can see that is common in both the terms in the expression. So we can take that out. From this we will get:
$ \Rightarrow k = {\left[ {a\left( {3{x^2} - 4ya{z^3}} \right)} \right]^2}$
Simplify it one step further, we’ll get:
$ \Rightarrow k = {a^2}{\left( {3{x^2} - 4ya{z^3}} \right)^2}$
Now, we will apply an algebraic formula of ${\left( {a - b} \right)^2}$ as show below:
$ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
By applying this formula in our expression, we’ll get:
$ \Rightarrow k = {a^2}\left[ {{{\left( {3{x^2}} \right)}^2} + {{\left( {4ya{z^3}} \right)}^2} - 2\left( {3{x^2}} \right)\left( {4ya{z^3}} \right)} \right]$
Simplifying it even further, we’ll get:
$
\Rightarrow k = {a^2}\left( {9{x^4} + 16{y^2}{a^2}{z^6} - 2 \times 3 \times 4{x^2}ya{z^3}} \right) \\
\Rightarrow k = {a^2}\left( {9{x^4} + 16{y^2}{a^2}{z^6} - 24{x^2}ya{z^3}} \right) \\
$
Thus this is the expanded form of our initial expression i.e. ${\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$.
$ \Rightarrow {\left( {3{x^2}a - 4ya{z^3}a} \right)^2} = {a^2}\left( {9{x^4} + 16{y^2}{a^2}{z^6} - 24{x^2}ya{z^3}} \right)$
We can write this in other forms also. For example:
$ \Rightarrow {\left( {3{x^2}a - 4ya{z^3}a} \right)^2} = {a^2}\left[ {9{x^4} + 8ya{z^3}\left( {2ya{z^3} - 24{x^2}} \right)} \right]$
We can write this many other different forms but its value will remain the same.
Note: Some of the important formulas that are widely used in the simplification of algebraic expressions are shown below:
$
\Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \\
\Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \\
\Rightarrow {a^3} + b{}^3 = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) \\
\Rightarrow {a^3} + b{}^3 = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) \\
\Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) \\
$
Out of these formulas, one which we have already used in the above question is \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]. Others also we can use for such simplifications wherever required.
Complete step by step answer:
According to the question, we have to show the process of simplification of an expression ${\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$.
Let this expression be denoted by a variable $k$. So we have:
$ \Rightarrow k = {\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$
We can see that is common in both the terms in the expression. So we can take that out. From this we will get:
$ \Rightarrow k = {\left[ {a\left( {3{x^2} - 4ya{z^3}} \right)} \right]^2}$
Simplify it one step further, we’ll get:
$ \Rightarrow k = {a^2}{\left( {3{x^2} - 4ya{z^3}} \right)^2}$
Now, we will apply an algebraic formula of ${\left( {a - b} \right)^2}$ as show below:
$ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
By applying this formula in our expression, we’ll get:
$ \Rightarrow k = {a^2}\left[ {{{\left( {3{x^2}} \right)}^2} + {{\left( {4ya{z^3}} \right)}^2} - 2\left( {3{x^2}} \right)\left( {4ya{z^3}} \right)} \right]$
Simplifying it even further, we’ll get:
$
\Rightarrow k = {a^2}\left( {9{x^4} + 16{y^2}{a^2}{z^6} - 2 \times 3 \times 4{x^2}ya{z^3}} \right) \\
\Rightarrow k = {a^2}\left( {9{x^4} + 16{y^2}{a^2}{z^6} - 24{x^2}ya{z^3}} \right) \\
$
Thus this is the expanded form of our initial expression i.e. ${\left( {3{x^2}a - 4ya{z^3}a} \right)^2}$.
$ \Rightarrow {\left( {3{x^2}a - 4ya{z^3}a} \right)^2} = {a^2}\left( {9{x^4} + 16{y^2}{a^2}{z^6} - 24{x^2}ya{z^3}} \right)$
We can write this in other forms also. For example:
$ \Rightarrow {\left( {3{x^2}a - 4ya{z^3}a} \right)^2} = {a^2}\left[ {9{x^4} + 8ya{z^3}\left( {2ya{z^3} - 24{x^2}} \right)} \right]$
We can write this many other different forms but its value will remain the same.
Note: Some of the important formulas that are widely used in the simplification of algebraic expressions are shown below:
$
\Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \\
\Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \\
\Rightarrow {a^3} + b{}^3 = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) \\
\Rightarrow {a^3} + b{}^3 = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) \\
\Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) \\
$
Out of these formulas, one which we have already used in the above question is \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]. Others also we can use for such simplifications wherever required.
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