99% Can't Solve This | A Very Nice Radical Problem | Can You Solve This ?

99% Can't Solve This | A Very Nice Radical Problem | Can You Solve This ?
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[@elecbertelecbert-l5e]

انت رائع

[@ManojkantSamal]

^=read as to the power
*=read as square root
Let, {(*5+1)/2}=R
{(*5-1)/2}=A
As per question
(R^16)-(A^16)=987.*5......eqn1
Let's explain,
(3+*5)/2
={(6+2.*5)/4
={(*5^2)+(1^2)+(2×*5×1)}/(2^2)
={(*5+1)^2}/(2^2)
={(*5+1)/2}^2
=R^2
So,
{(3+*5)/2}^16=(R^2)^16=R^32
Similarly we can get
{(3-*5)/2}^16=A^32
As per question
(R^32)+(A^32)=?
Now
RA={(*5+1)(*5-1)}/(2×2)
=(5-1)/4=4/4=1....eqn2
(RA)^16=1^16=1
(R^16).(A^16)=1...eqn3
Now take the square of eqn1
(R^16)^2+(A^16)^2-{2.R^16).(A^16)}=(987.*5)^2
So,
(R^32)+(A^32)-2=4870845
So,
(R^32)+(A^32)=4870845+2
=4870847
Hence,
{(3+*5)/2}^16 +{(3-*5)/2}^16=4870847....…. May be

[@raghvendrasingh1289]

❤

[@asvquickcalculations6712]

Let (3+root5)/2=x
3+root5=2x
root5=,[2x-3.] root5+1=2x-2 androot5-1=2x-4
root(5+1)/2=x-1 ( root5-1)/2=x-2

[@RealQinnMalloryu4]

(2^3+1/1)2^4 ➖ (2^3 ➖ 1/1)2^4 (1^1+/)2^2^2 ➖ ( ➖ 1^3/)^2^2^2 (+/)^1^1^1 ➖ (3/)1^1^2.(3/)^2 (x ➖ 3x+2). (1+2^3/1)^2^4 ➖ (1 ➖ 2^3/1)^2^4 (+1^1/)^1^1^1 ➖ ( ➖ 1^3)^1^1^2 (3/)^2 (x ➖ 3x+2).