Let 2000 times the cube of x be equal to 200 1 times the cube of y be equal to 2002 times the cube of z be equal to k 3 2000 x? =K^3 /x 200 1y？ =K^3 /y 2002z？ =K^3 /z

3 √( 2000)x = 3 √( 2006 54 38+0)y = 3 √( 2002)z = k

3 √( 2000)+3 √( 2006 54 38+0)+3 √( 2002)= k( 1÷x+ 1÷y+ 1÷z)= 3 √( 2000 x？ +200 1y？ +2002z？ )=3√[k^3( 1 \u x+ 1 \u y+ 1 \u z)]

X times y times z is greater than zero.

( 1 \u x+ 1 \u y+ 1 \u z)= 3 √( 1 \u x+ 1 \u y+ 1 \u z)

( 1 \u x+ 1 \u y+ 1 \u z)= 1

You got it?