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Hint:The given problem is related to simplification by factorization. Express each term as a product of its factors and then simplify the expression.

Complete step-by-step answer:

We will proceed with the solution by taking each term, factoring it, and then substituting the factored form in the given expression. Then we will simplify the given expression.

The given expression is $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$. The first term of the expression is ${{a}^{2}}+10a+21$. It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $10a$ into two terms, such as their sum is equal to $10a$ and their product is equal to $21{{a}^{2}}$ .

We can write $10a$ as $7a+3a$ . Here, the sum of $7a$ and $3a$ is $10a$ and the product of $7a$ and $3a$ is $21{{a}^{2}}$ . So, ${{a}^{2}}+10a+21$can be written as ${{a}^{2}}+7a+3a+21$ .

$\Rightarrow {{a}^{2}}+10a+21=a\left( a+7 \right)+3\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+10a+21=\left( a+3 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+10a+21$ as $\left( a+3 \right)\left( a+7 \right)$ .

Now, the second term is ${{a}^{2}}+6a-7$ . It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $6a$ into two terms, such as their sum is equal to $6a$ and their product is equal to \[-7{{a}^{2}}\] .

We can write $6a$ as $7a-a$ . Here, the sum of $7a$ and $-a$ is $6a$ and the product of $7a$ and $-a$ is $-7{{a}^{2}}$ . So, ${{a}^{2}}+6a-7$can be written as ${{a}^{2}}+7a-a-7$ .

$\Rightarrow {{a}^{2}}+6a-7=a\left( a+7 \right)-1\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+6a-7=\left( a-1 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+6a-7$ as $\left( a-1 \right)\left( a+7 \right)$ .

Now, the third term is ${{a}^{2}}-1$ . We know, we can write ${{a}^{2}}-1$ as ${{a}^{2}}-{{1}^{2}}$. So, ${{a}^{2}}-1={{a}^{2}}-{{1}^{2}}=\left( a+1 \right)\left( a-1 \right)$ . So, we can write ${{a}^{2}}-1$ as $\left( a+1 \right)\left( a-1 \right)$

Now, the fourth term is $\left( a+3 \right)$ . It is already in its simplest form. So, we don’t need to factorize it.

Now, the expression $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ can be written as $\dfrac{\left( a+7 \right)\left( a+3 \right)}{\left( a+7 \right)\left( a-1 \right)}\times \dfrac{\left( a-1 \right)\left( a+1 \right)}{\left( a+3 \right)}$ .

$\Rightarrow \dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}=a+1$

Hence, the simplified value of $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ is $a+1$ .

Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

Complete step-by-step answer:

We will proceed with the solution by taking each term, factoring it, and then substituting the factored form in the given expression. Then we will simplify the given expression.

The given expression is $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$. The first term of the expression is ${{a}^{2}}+10a+21$. It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $10a$ into two terms, such as their sum is equal to $10a$ and their product is equal to $21{{a}^{2}}$ .

We can write $10a$ as $7a+3a$ . Here, the sum of $7a$ and $3a$ is $10a$ and the product of $7a$ and $3a$ is $21{{a}^{2}}$ . So, ${{a}^{2}}+10a+21$can be written as ${{a}^{2}}+7a+3a+21$ .

$\Rightarrow {{a}^{2}}+10a+21=a\left( a+7 \right)+3\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+10a+21=\left( a+3 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+10a+21$ as $\left( a+3 \right)\left( a+7 \right)$ .

Now, the second term is ${{a}^{2}}+6a-7$ . It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $6a$ into two terms, such as their sum is equal to $6a$ and their product is equal to \[-7{{a}^{2}}\] .

We can write $6a$ as $7a-a$ . Here, the sum of $7a$ and $-a$ is $6a$ and the product of $7a$ and $-a$ is $-7{{a}^{2}}$ . So, ${{a}^{2}}+6a-7$can be written as ${{a}^{2}}+7a-a-7$ .

$\Rightarrow {{a}^{2}}+6a-7=a\left( a+7 \right)-1\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+6a-7=\left( a-1 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+6a-7$ as $\left( a-1 \right)\left( a+7 \right)$ .

Now, the third term is ${{a}^{2}}-1$ . We know, we can write ${{a}^{2}}-1$ as ${{a}^{2}}-{{1}^{2}}$. So, ${{a}^{2}}-1={{a}^{2}}-{{1}^{2}}=\left( a+1 \right)\left( a-1 \right)$ . So, we can write ${{a}^{2}}-1$ as $\left( a+1 \right)\left( a-1 \right)$

Now, the fourth term is $\left( a+3 \right)$ . It is already in its simplest form. So, we don’t need to factorize it.

Now, the expression $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ can be written as $\dfrac{\left( a+7 \right)\left( a+3 \right)}{\left( a+7 \right)\left( a-1 \right)}\times \dfrac{\left( a-1 \right)\left( a+1 \right)}{\left( a+3 \right)}$ .

$\Rightarrow \dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}=a+1$

Hence, the simplified value of $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ is $a+1$ .

Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

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