Question

Find a value of x such that :


( 3+ 4) (3^2 + 4^2) ( 3^4 + 4^4)(3^8 + 4^8)
( 3^16 + 4^16 ) ( 3^32 + 4^32 ) = 4^x - 3^x


​

Answer 1

→ Hey Mate,

→ Given Question : -

→  $( 3 + 4 ) ( 3^2 + 4 ^2 ) ( 3^4 + 4^4 ) ( 3^8 +4^8) (3^1^6 + 4^1^6 ) (3^3^2 + 4 ^3^2 )$

→  $4^x -3^x$

→ Formula Used :-

→ $( a - b) (a + b) =a^2 -b^2$

→ Solution :-

→ Step 1 : reverse  the order of additional in each binomial.

→  $( 4 + 3 ) ( 4^2 + 3 ^2 ) ( 4^4 +3^4 ) (4^8 + 3 ^8) ( 4^1^6 + 3^1^6)( 4^3^2 + 3^3^2)$

→  $4^x -3^x$

→ step 2 : multiply both sides by ( 4 - 3 ) = 1

→ $( 4 - 3) ( 4 + 3 ) ( 4^2 +3^2 ) ( 4^4 + 3^4 ) ( 4^8 + 3^8) ( 4 ^1^6 + 3 ^ 1^6) ( 4^3^2+3^3^2)$

→ $( 4^x -3^x ) ( 4 -3 )$

→ step 3 : notice the product is a different of squares .

→ $( 4-3 ) ( 4 +3 ) (4^2 +3^2 ) (4^4 + 3^4 ) ( 4^8 +3^8 ) ( 4^1^6 +3^1^6 ) ( 4^3^2 + 3^3^2)$ =

→ $4^x -3^x$

→ $( 4^2 - 3^2 ) (4^2 + 3^2 ) (4^4 +3^4 ) ( 4^8 +3^8 ) ( 4^1^6 + 3^1^6 ) ( 4^3^2 + 3^3^2 )$

→ $4^x -3^x$

→ $( 4^4 - 3^4 ) ( 4^4 + 3^4 ) (4 ^8 + 3^8 ) ( 4^1^6 + 3^1^6 ) ( 4^3^2 + 3^3^2 )$

→ $4^x - 3^x$

→$( 4 ^8 -3^8 ) ( 4^8 + 3^8 ) ( 4^1^ 6 + 3^1^6) ( 4^3^2 + 3^3^2 )$

→ $4^x -3^x$

→ $( 4^1^6 - 3^1^6 ) ( 4^1^6 + 3^1^6 ) ( 4^3^2 + 4^3^2 )$

→ $4^x - 3^x$

→$( 4^3^2 - 3^3 ^2 ) ( 4^3^2 +3^3^2 ) = 4^x - 3^x$

→$( 4^6^4 - 3^6^4 ) = 4^x - 3^x$

→ so,  the value of x = 64 is in the above solution.

-------------------------------------------------------------------------------------------------------------

Answer 2

Step-by-step explanation:

Myself Gaurav (Real name)

Will u be my friend???

Hmm