Expand the following expressions:
\[\left( i \right)\left( 3{{x}^{4}}-5 \right)\left( 7{{x}^{5}}-2x \right)\]
\[\left( ii \right){{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}\]

Hint: To solve the given question, we will open both the given brackets using the rule of multiplication given as \[\left( a+b \right)\left( x+y \right)=ax+ay+bx+by.\] Using this, we will assume \[a=3{{x}^{4}},b=5,etc.\] for all the terms given in (i) and (ii) and hence solve accordingly.

Complete step by step answer:

The multiplication of expansion of two given brackets product is given by the formula
\[\left( a+b \right)\left( x+y \right)=a\left( x+y \right)+b\left( x+y \right)\]
\[\Rightarrow \left( a+b \right)\left( x+y \right)=ax+ay+bx+by\]
So, we will use this technique in both (i) and (ii) to solve further.
\[\left( i \right)\left( 3{{x}^{4}}-5 \right)\left( 7{{x}^{5}}-2x \right)\]
Using the formula and process stated above, we have,
\[\Rightarrow \left( 3{{x}^{4}}-5 \right)\left( 7{{x}^{5}}-2x \right)=3{{x}^{4+5}}7-6{{x}^{4+1}}-35{{x}^{5}}+10x\]
\[\Rightarrow \left( 3{{x}^{4}}-5 \right)\left( 7{{x}^{5}}-2x \right)=21{{x}^{9}}-6{{x}^{5}}-35{{x}^{5}}+10x\]
\[\Rightarrow \left( 3{{x}^{4}}-5 \right)\left( 7{{x}^{5}}-2x \right)=21{{x}^{9}}-41{{x}^{5}}+10x\]
So, the value of \[\left( i \right)\left( 3{{x}^{4}}-5 \right)\left( 7{{x}^{5}}-2x \right)=21{{x}^{9}}-41{{x}^{5}}+10x.\]
\[\left( ii \right){{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}\]
To solve this part, we will separately solve \[{{\left( 2x-7 \right)}^{2}}\] and \[{{\left( 3x+5 \right)}^{3}}.\]
The formula of \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] and of \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right).\]
So, using the above stated formula of \[{{\left( a-b \right)}^{2}}\] on \[{{\left( 2x-7 \right)}^{2}}\] we get,
\[{{\left( 2x-7 \right)}^{2}}={{\left( 2x \right)}^{2}}-2\times 2x\times 7+{{\left( 7 \right)}^{2}}\]
\[\Rightarrow {{\left( 2x-7 \right)}^{2}}=4{{x}^{2}}-28x+49\]
Again using the above stated formula of \[{{\left( a+b \right)}^{3}}\] in \[{{\left( 3x+5 \right)}^{3}}\] we get,
\[{{\left( 3x+5 \right)}^{3}}={{\left( 3x \right)}^{3}}+{{5}^{3}}+3\times 3x\times 5\left( 3x+5 \right)\]
\[\Rightarrow {{\left( 3x+5 \right)}^{3}}=27{{x}^{3}}+125+45x\left( 3x+5 \right)\]
\[\Rightarrow {{\left( 3x+5 \right)}^{3}}=27{{x}^{3}}+125+45x\times 3{{x}^{2}}+45\times 5x\]
\[\Rightarrow {{\left( 3x+5 \right)}^{3}}=27{{x}^{3}}+125+135{{x}^{2}}+225x\]
So, we finally have,
\[{{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}=\left( 4{{x}^{2}}-28x+49 \right)\left( 27{{x}^{3}}+125+135{{x}^{2}}+225x \right)\]
Using, \[{{x}^{m}}.{{x}^{n}}={{x}^{m+n}},\] we get,
\[\begin{align}
  & \Rightarrow {{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}=4\times 27{{x}^{5}}+4\times 125{{x}^{2}}+4\times 135{{x}^{4}}+225\times 4{{x}^{3}}-28\times 27{{x}^{4}}-28\times 125x \\
 & -28\times 135{{x}^{3}}-28\times 225{{x}^{2}}+49\,\times 27{{x}^{3}}+49\times 125+49\times 135{{x}^{2}}+49\times 225x \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}=108{{x}^{5}}+500{{x}^{2}}+540{{x}^{4}}+900{{x}^{3}}-756{{x}^{4}}-3500x-3780{{x}^{3}} \\
 & -6300{{x}^{2}}+1323{{x}^{3}}+6125+6615{{x}^{2}}+11025x \\
\end{align}\]
\[\Rightarrow {{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}=108{{x}^{5}}-216{{x}^{4}}+815{{x}^{2}}+7525x-1557{{x}^{3}}+6125\]
Therefore, we get
\[\Rightarrow {{\left( 2x-7 \right)}^{2}}{{\left( 3x+5 \right)}^{3}}=108{{x}^{5}}-216{{x}^{4}}-1557{{x}^{3}}+815{{x}^{2}}+7525x+6125\]


Note:
Another way to solve the part \[{{\left( 3x+5 \right)}^{3}}\] can be directed by using \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}.\] We have used the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right).\] Both of the above formulas are the same and hence the answer would be the same. One key point to note here in this question is that while multiplying \[\left( 4{{x}^{2}}-28x+49 \right)\] to \[\left( 27{{x}^{3}}+125+135{{x}^{2}}+225x \right)\] the minus in “– 28” should be taken care of while multiplying.