Question

The value of (a ^b+b^a)^-1 when a=2 and b=3 is

Answer 1

Answer:

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Step-by-step explanation:

We have

ab+ba=1⟹a2+b2−abab=0

Now back to the right hand side

a3+b3=(a+b)(a2−ab+b2)

We can notice that we have

a2+b2−abab=0⟹a2+b2−ab=0

Trivially a,b≠0 .

Thus we have

a3+b3=(a+b)(a2−ab+b2)=(a+b)⋅0=0

The result is

a3+b3=0

Answer 2

Answer:

1/17

Step-by-step explanation:

(a^b+b^a)^-1

(2³+3²)^-1

(8+9)^-1

(17)^-1

1/17

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