Question

What is the value of 77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3?

Answer 1

Solution:-  

Assume  

a =  77! And  

b=54!  

according to the expression    

(a(a-2b)^3)/(a+b)^3 + (b(2a-b)^3)/(a+b)^3)

=(a^4+2×a^3×b -2×a×b^3-b^4)/(a+b)^3      

= (a-b)×(a+b)^3/(a+b)^3      

= (a-b)    

= (77!-54!)      

= 1.45183092e113 Ans  

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@GauravSaxena01

Answer 2

Answer:

$\Rightarrow \frac{77!(77!-2(54!))^3}{(77!+54!)^3}+\frac{54!(2(77!)-54!)^3}{(77!+54!)^3}=8.5\times 10^{101}$

Explanation:

$\text{Given expression: }\frac{77!(77!-2(54!))^3}{(77!+54!)^3}+\frac{54!(2(77!)-54!)^3}{(77!+54!)^3}$

Let us suppose 77! = a and 54! = b

Substitute above assumption into expression and we get

$\Rightarrow \frac{a(a-2b)^3}{(a+b)^3}+\frac{b(2a-b)^3}{(a+b)^3}$

Make common denominator

$\Rightarrow \frac{a(a-2b)^3+b(2a-b)^3}{(a+b)^3}$

Formula:

$(x-y)^3=x^3-3x^2y+3xy^2-y^3$

$(x+y)^3=x^3+3x^2y+3xy^2+y^3$

Apply the formula in given expression and we get

$\Rightarrow \frac{a(a^3-6a^2b+12ab^2-8b^3)+b(8a^3-12a^2b+6ab^2-b^3)}{(a+b)^3}$

Simplify using distributive property and Combine the like terms

$\Rightarrow \frac{a^4+2a^3b-2ab^3-b^4}{(a+b)^3}$

Factor numerator and simplify it

$\Rightarrow \frac{(a+b)^3(a-b)}{(a+b)^3}$

Simplify the fraction by canceling like factors from numerator and denominators

$\Rightarrow a-b$

where, a=71! and b=54!

$\Rightarrow 71!-54!$

using calculator ,

$\Rightarrow 71!-54!=8.5\times 10^{101}$