The solution to the equation \[1 + a + {a^2} + {a^3} + \cdots + {a^x} = \left( {1 + a} \right)\left( {1 + {a^2}} \right)\left( {1 + {a^4}} \right)\]. What is the value of \[x\]?
A. 3
B. 5
C. 7
D. None of these

Hint First we will simplify the left side expression of the equation by using the sum of geometric series. Then solve the equation and compare the power of \[a\].

Formula Used:
The sum of geometric terms is \[S = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\], \[r \ne 1\] and \[r < 1\]
Or, \[S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\], \[r \ne 1\] and \[r > 1\]

Complete step by step solution
Given: \[1 + a + {a^2} + {a^3} + \cdots + {a^x} = \left( {1 + a} \right)\left( {1 + {a^2}} \right)\left( {1 + {a^4}} \right)\]
Where \[a{\rm{ }} = {\rm{ }}1\] ,\[r{\rm{ }} = {\rm{ }}a\], \[n{\rm{ }} = {\rm{ }}x + 1\],
\[ \Rightarrow \dfrac{{1(1 - {a^{x + 1}})}}{{(1 - a)}} = {\rm{ }}\left( {1 + a} \right)(1 + {a^2})(1 + {a^4})\]
Send the Denominator or LHS to RHS,
\[ \Rightarrow 1(1 - {a^{x + 1}}) = (1 - a)\left( {1 + a} \right)(1 + {a^2})(1 + {a^4})\]
We have used the formula, \[(a - b)(a + b) = {a^2} - {b^2}\] \[\]
 \[ \Rightarrow 1(1 - {a^{x + 1}}) = (1 - {a^2})(1 + {a^2})(1 + {a^4})\]
Similarly, we have used the same formula here,
 \[ \Rightarrow 1(1 - {a^{x + 1}}) = (1 - {a^4})(1 + {a^4})\]
And like before, \[(a - b)(a + b) = {a^2} - {b^2}\]has been used here to get the below equation,
 \[ \Rightarrow 1(1 - {a^{x + 1}}) = (1 - {a^8})\]
Now as LHS and RHS are similar in expression, we compare the exponents to get
 \[x + 1 = 8\]
 \[x = 7\]
Hence, the correct option is option C.

Note: To solve this question, you need to know the formula sum of geometric sequence. Here you need to apply the sum of the geometric sequence on the left side of the equation. Since the first term of the sequence is 1 and the last term \[{a^x}\], the total number of terms is \[x + 1\]. Assume that the value of \[a < 1\]. Apply the formula \[S = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\] and solve the equation.