Compute: \[\dfrac{{({{10}^4} + 324)({{22}^4} + 324)({{34}^4} + 324)({{46}^4} + 324)({{58}^4} + 324)}}{{({4^4} + 324)({{16}^4} + 324)({{28}^4} + 324)({{40}^4} + 324)({{52}^4} + 324)}}\]
Answer
277.5k+ views
Hint: The question is in the form of exponential. It is in the form of \[\left( {{{\left( {{a^2}} \right)}^2} + {b^2}} \right)\], The b term for each and every term is same. The value of a is different for other terms. The given terms are in the form of \[({x^4} + 324)\], this can be written as \[({(x + 3)^2} + 9)({(x - 3)^2} + 9)\], this formula is implemented to each term and then we are going to simplify.
Complete step by step answer:
Consider the term \[({(x + 3)^2} + 9)({(x - 3)^2} + 9)\], on multiplying this we have
\[ \Rightarrow {(x + 3)^2}({(x - 3)^2} + 9) + 9({(x - 3)^2} + 9)\]
Multiplying to each term we have
\[ \Rightarrow {(x + 3)^2}{(x - 3)^2} + 9{(x + 3)^2} + 9{(x - 3)^2} + 81\]
On simplifying we have
\[ \Rightarrow {({x^2} - 9)^2} + 9(2{x^2} + 18) + 81\]
Expanding the terms
\[ \Rightarrow {x^4} + 81 - 18{x^2} + 18{x^2} + 162 + 81\]
On further simplification we have
\[ \Rightarrow {x^4} + 324\]
Therefore \[({x^4} + 324) = ({(x - 3)^2} + 9)({(x + 3)^2} + 9)\]----------- (1)
Now consider the given question
\[ \Rightarrow \dfrac{{({{10}^4} + 324)({{22}^4} + 324)({{34}^4} + 324)({{46}^4} + 324)({{58}^4} + 324)}}{{({4^4} + 324)({{16}^4} + 324)({{28}^4} + 324)({{40}^4} + 324)({{52}^4} + 324)}}\]
By using the equation (1) the above inequality is written as
\[ \Rightarrow \dfrac{{({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)({{61}^2} + 9)}}{{({1^2} + 9)({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)}}\]
When we see both the numerator and the denominator there are so many common terms hence we can cancel them.
On cancelling them we have
\[ \Rightarrow \dfrac{{({{61}^2} + 9)}}{{({1^2} + 9)}}\]
Now we have a very simplified term, in the numerator there is only one term and in the denominator also we can see only one term.
The terms which are in the form of exponents. On expanding it we have
\[ \Rightarrow \dfrac{{(61 \times 61 + 9)}}{{(1 \times 1 + 9)}}\]
Now multiply 61 and 61 in the numerator. Multiply 1 and 1 in the denominator.
\[ \Rightarrow \dfrac{{(3721 + 9)}}{{(1 + 9)}}\]
Now add 3721 and 9 in the numerator. Add 1 and 9 in the denominator.
\[ \Rightarrow \dfrac{{3730}}{{10}}\]
On dividing the 3730 by 10
\[ \Rightarrow 373\]
Hence we have simplified the given inequality.
Note: The number can be written in the form of the exponents and the exponent number can also be written in the form of the usual number. We must know about the law of exponents. Since the given terms are exponential, we need a law of exponents to solve.
Complete step by step answer:
Consider the term \[({(x + 3)^2} + 9)({(x - 3)^2} + 9)\], on multiplying this we have
\[ \Rightarrow {(x + 3)^2}({(x - 3)^2} + 9) + 9({(x - 3)^2} + 9)\]
Multiplying to each term we have
\[ \Rightarrow {(x + 3)^2}{(x - 3)^2} + 9{(x + 3)^2} + 9{(x - 3)^2} + 81\]
On simplifying we have
\[ \Rightarrow {({x^2} - 9)^2} + 9(2{x^2} + 18) + 81\]
Expanding the terms
\[ \Rightarrow {x^4} + 81 - 18{x^2} + 18{x^2} + 162 + 81\]
On further simplification we have
\[ \Rightarrow {x^4} + 324\]
Therefore \[({x^4} + 324) = ({(x - 3)^2} + 9)({(x + 3)^2} + 9)\]----------- (1)
Now consider the given question
\[ \Rightarrow \dfrac{{({{10}^4} + 324)({{22}^4} + 324)({{34}^4} + 324)({{46}^4} + 324)({{58}^4} + 324)}}{{({4^4} + 324)({{16}^4} + 324)({{28}^4} + 324)({{40}^4} + 324)({{52}^4} + 324)}}\]
By using the equation (1) the above inequality is written as
\[ \Rightarrow \dfrac{{({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)({{61}^2} + 9)}}{{({1^2} + 9)({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)}}\]
When we see both the numerator and the denominator there are so many common terms hence we can cancel them.
On cancelling them we have
\[ \Rightarrow \dfrac{{({{61}^2} + 9)}}{{({1^2} + 9)}}\]
Now we have a very simplified term, in the numerator there is only one term and in the denominator also we can see only one term.
The terms which are in the form of exponents. On expanding it we have
\[ \Rightarrow \dfrac{{(61 \times 61 + 9)}}{{(1 \times 1 + 9)}}\]
Now multiply 61 and 61 in the numerator. Multiply 1 and 1 in the denominator.
\[ \Rightarrow \dfrac{{(3721 + 9)}}{{(1 + 9)}}\]
Now add 3721 and 9 in the numerator. Add 1 and 9 in the denominator.
\[ \Rightarrow \dfrac{{3730}}{{10}}\]
On dividing the 3730 by 10
\[ \Rightarrow 373\]
Hence we have simplified the given inequality.
Note: The number can be written in the form of the exponents and the exponent number can also be written in the form of the usual number. We must know about the law of exponents. Since the given terms are exponential, we need a law of exponents to solve.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
