Question

help me to solve this math problem. here you have to prove Right hand side is equal to x^6​

Answer 1

Answer:

Step-by-step explanation:

Given that,

p + q + r = 3 and pq + qr + pr = 3.

We know that,

( p + q + r )² = p² + q² + r² + 2 ( pq + qr + pr )

( 3 )² = p² +q² + r² + 2 (3 )

9 = p² + q² + r² + 6

p² + q² + r² = 9 - 6

p² + q² + r² = 3 .

( xy )^ ( p - 1 ) ( p + 1 ) × ( xy )^ ( q - 1 ) ( q + 1 ) × ( xy )^ ( r - 1 ) ( r + 1 )

(" ^ " this symbol means " to the power of " )

= ( xy )^( p² - 1 ) × ( xy )^(q² - 1 ) × ( xy )^(r²- 1 ) [∵(a - b) ( a + b) = a² -b² ]

= ( xy ) ^( p² - 1 + q² - 1 + r² - 1 ) (∵Bases are equal, then powers should be add )

= ( xy ) ^ (p² + q² + r² - 3 )

= ( xy ) ^ ( 3 - 3 ) ( ∵ p² + q² + r² = 3 )

= ( xy )^0 ( ∵ a^0 = 1 )

= 1 is the answer.