$$\displaystyle \left\{ \left( { 2b }^{ 2 }-2bc-{ c }^{ 2 } \right) +\left( { c }^{ 2 }+2bc-{ b }^{ 2 } \right)  \right\} -\left\{ \left( { 9b }^{ 2 }+{ 3c }^{ 2 } \right) +\left( { 2b }^{ 2 }+bc + { 2c }^{ 2 } \right)  \right\} $$
=$$\displaystyle \left\{ \left( { 2b }^{ 2 }-{ b }^{ 2 } \right) +\left( -2bc+2bc \right) +\left( { c }^{ 2 }-{ c }^{ 2 } \right)  \right\} -\left\{ \left( { 9b }^{ 2 }+{ 2b }^{ 2 } \right) +\left( { 3c }^{ 2 }+{ 2c }^{ 2 } \right) +bc \right\} $$
$$\displaystyle =\left( { b }^{ 2 } \right) -\left( { 11b }^{ 2 }+{ 5c }^{ 2 }+bc \right) $$

$$\displaystyle ={ b }^{ 2 }-{ 11b }^{ 2 }-{ 5c }^{ 2 }-bc$$

$$\displaystyle =-{ 10b }^{ 2 }-bc-{ 5c }^{ 2 }$$.

$$=ad-bc+bc-ca+ca-ab$$

$$=0$$

$$(a^4-3a^2+6a-8)-(a^4-10a+4a^2+8)$$

$$=a^4-3a^2+6a-8-a^4+10a-4a^2-8$$

$$=a^4-a^4-3a^2-4a^2+6a+10a-8-8$$

$$=-7a^2+16a-16$$

So,
$$(x^3 + 3x - 8) +Y= 3x^3 + x^2 + 6$$
$$\Rightarrow Y=(3x^3 + x^2 + 6) -(x^3 + 3x - 8)$$
$$\Rightarrow Y=3x^3 + x^2 + 6 -x^3 - 3x + 8$$
$$\Rightarrow Y=2x^3+x^2-3x+14$$

So, option $$A$$ is correct.

$$(2b^3c^2+b^2c-4bc)-(b^3c^2-b^2c-4bc)$$

$$ (2x^2+4x+8)-(2x^2-4x+7)$$

$$= (5jk^2 - jk^2) + (5j^2 - 5j^2) + (-2j^2k - 2j^2k)$$

$$= 4jk^2 - 4j^2k=-(4j^2k-4k^2j)$$

$$\Rightarrow 8x+14-6x+12$$

$$\Rightarrow 2x+26$$

$$(x^2y-3y^2+5xy^2)-(-x^2y+3xy^2-3y^2) $$

$$\Rightarrow t^2-82t^2+60t-59t+54$$

$$\Rightarrow -81t^2+t+54$$

$$\Rightarrow a+2b+3c-4a-6b+5c$$

$$\Rightarrow -3a-4b+8c$$

$$\left ( -3pq^{2}+4pq \right )-\left ( 7pq-9pq^{2} \right )$$

$$=-3pq^{2}+4pq-7pq+9pq^{2}$$

$$=-3pq^{2}+9pq^{2}+4pq-7pq$$

$$=6pq^{2}-3pq$$

$$(-8pst+3prq-14)-(7prq-20+2pst)$$

$$=-8pst+3prq-14-7prq+20-2pst$$

$$=-8pst-2pst+3prq-7prq-14+20$$

$$=-10pst-4prq+6$$

$$=6-10pst-4prq$$

Expression be x

$$\Rightarrow x+(2a^2+3a+2)+(3a^2-4a+1)$$$$=8+13a+7a^2$$

$$=x+5a^2-a+3$$$$=8+13a+7a^2$$

$$\Rightarrow x=7a^2+13a+8-5a^2+a-3$$

$$x=2a^2+14a+5$$.