Question

If x 3^a, y = 9^b, z = 27^c और x^bcy^ca
z^ab = 1 तो abc = |
(a) 1
(c) 27
(d) 0
(b) 27

Answer 1

Step-by-step explanation:

$x = 3 {}^{a} \\ y = 9 {}^{b} \\ z = 27 {}^{c} \\ and \\ x {}^{bc} \times y {}^{ca} \times z {}^{ab} = 1 \\( 3 {}^{a} ) {}^{bc} \times (9 {}^{b} ) {}^{ca} \times (27 {}^{c} ) {}^{ab} = 1 \\ 3 {}^{abc} \times 3 {}^{2abc} \times 3 {}^{3abc} = 1 \\ 3 {}^{abc + 2abc + 3abc} = 3 {}^{0} \\ 6abc = 0 \\ abc = 0$

Answer 2

Answer:

Step-by-step explanation: