Question

(64/125)^-2/3 + 1/(256/625)^1/4 +(root 25/ 3 root 64)^0=61/16​

Answer 1

Step-by-step explanation:

(64/125)^(-2/3) + 1/(256/625)^(1/4) + (root 25/3root64)^0 =61/16

Taking LHS

(64/125)^(-2/3) + 1/(256/625)^(1/4) + (root25/ 3root64)^0

as x^-y = 1/x^y

and x^0 = 1

then above equation will be

(125/64)^(2/3) + 1/[(4/5)^4]^(1/4) + 1

[(5/4)^3]^(2/3) + 1/(4/5)^1 + 1

(5/4)^2 + 5/4 +1

25/16 + 5/4 +1

$\frac{25 + 20 + 16}{16}$

61/16 = RHS

HENCE PROVED

HOPE IT HELPS YOU