4 years ago

If P=Q^R , Q=R^P , R=P^Q.Then Prove: P×Q×R=1.

Mathematics

1 answer:

8_murik_8 [283]4 years ago

5 0

Answer:

PQR=1

Step-by-step explanation:

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3 years ago

If a,b,c,d, and e are non negative such that x^6-289x^4-x^2+289=(x-a)(x+b)(x_c)(x+d)(x^2+ex+1), what is the value of a+b+c+d+e?

Irina18 [472]

Answer:

Step-by-step explanation:

Hello,

If $X=x^2$ it becomes

$X^3-289X^2-X+289=(X^2-1)(X-289)$

and $289=17^2$

$X^2-1=(X-1)(X+1)=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1) \\\\x^2-289=x^2-17^2=(x-17)(x+17)$

So,

$x^6-289x^4-x^2+289=(x^2-1)(x^2+1)(x^2-289)=(x-1)(x+1)(x+17)(x-17)(x^2+1)$

a = 1

b = 1

c = 17

d = 17

e = 0

Then a + b + c + d + e = 36

Thanks

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3 years ago

If P=Q^R , Q=R^P , R=P^Q.Then Prove: P×Q×R=1.​

If P=Q^R , Q=R^P , R=P^Q.Then Prove: P×Q×R=1.