
If $ {x^4} + \dfrac{1}{{{x^4}}} = 194 $ , then $ {x^3} + \dfrac{1}{{{x^3}}} $ is equal to
$ \left( a \right){\text{ 76}} $
$ \left( b \right){\text{ 52}} $
$ \left( c \right){\text{ 64}} $
$ \left( d \right){\text{ None of these}} $
Answer
565.2k+ views
Hint: For solving this type of question we always have to frame the equation in such a way that it will follow the formula which is given by $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $ . In this we will add $ 2 $ both the sides and framing the equation, we get the value for $ {x^3} + \dfrac{1}{{{x^3}}} $ .
Formula used:
The algebraic formula will be given by
$ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
Here, $ a\& b $ will be the variables.
Complete step-by-step answer:
We have the equation given as $ {x^4} + \dfrac{1}{{{x^4}}} = 194 $
Now on adding both the sides $ 2 $ , we get the equation as
$ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = 194 + 2 $
Now by using the formula, the left side of the equation will be given as
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = 196 $
By removing the square from the left sides of the equation, we will get the equation as
$ \Rightarrow \left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 14 $
Now again adding $ 2 $ both the sides of the equation, we will get
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 14 + 2 $
Again we can see that the left side of the equation is following the formula $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = 16 $
By removing the square from the left sides of the equation, we will get the equation as
$ \Rightarrow x + \dfrac{1}{x} = 4 $ , and we will name it equation $ 1 $
Now for finding the $ {x^3} + \dfrac{1}{{{x^3}}} $ , we will take out the cube root of the equation $ 1 $ .
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {4^3} $
We get,
\[ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x + \dfrac{1}{x}} \right)\]
And from this, we will get
$ \Rightarrow {4^3} = {x^3} + \dfrac{1}{{{x^3}}} +3 \times 4 $
And on solving and taking the constant term to one side, we get the equation as
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 52 $
Hence, the option $ \left( b \right) $ is correct.
So, the correct answer is “Option b”.
Note: Here, in this question, we can see that with the help of the formula we can easily solve this type of question. So the important thing for us will be to memorize the formula and by practice, we will get the knowledge of substituting the formula with the question. Also while solving the question for the long answer question, we should follow each step.
Formula used:
The algebraic formula will be given by
$ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
Here, $ a\& b $ will be the variables.
Complete step-by-step answer:
We have the equation given as $ {x^4} + \dfrac{1}{{{x^4}}} = 194 $
Now on adding both the sides $ 2 $ , we get the equation as
$ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = 194 + 2 $
Now by using the formula, the left side of the equation will be given as
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = 196 $
By removing the square from the left sides of the equation, we will get the equation as
$ \Rightarrow \left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 14 $
Now again adding $ 2 $ both the sides of the equation, we will get
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 14 + 2 $
Again we can see that the left side of the equation is following the formula $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = 16 $
By removing the square from the left sides of the equation, we will get the equation as
$ \Rightarrow x + \dfrac{1}{x} = 4 $ , and we will name it equation $ 1 $
Now for finding the $ {x^3} + \dfrac{1}{{{x^3}}} $ , we will take out the cube root of the equation $ 1 $ .
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {4^3} $
We get,
\[ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x + \dfrac{1}{x}} \right)\]
And from this, we will get
$ \Rightarrow {4^3} = {x^3} + \dfrac{1}{{{x^3}}} +3 \times 4 $
And on solving and taking the constant term to one side, we get the equation as
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 52 $
Hence, the option $ \left( b \right) $ is correct.
So, the correct answer is “Option b”.
Note: Here, in this question, we can see that with the help of the formula we can easily solve this type of question. So the important thing for us will be to memorize the formula and by practice, we will get the knowledge of substituting the formula with the question. Also while solving the question for the long answer question, we should follow each step.
Recently Updated Pages
Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

The coating formed on the metals such as iron silver class 12 chemistry CBSE

Metals are refined by using different methods Which class 12 chemistry CBSE

What do you understand by denaturation of proteins class 12 chemistry CBSE

Assertion Nitrobenzene is used as a solvent in FriedelCrafts class 12 chemistry CBSE

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

What are the 12 elements of nature class 8 chemistry CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Full form of STD, ISD and PCO

Convert 40circ C to Fahrenheit A 104circ F B 107circ class 8 maths CBSE

Advantages and disadvantages of science


