Question

(3-x)^5+(x-5)^5=-32
what are the roots of this equation​

Answer 1

Step-by-step explanation:

$(3-x)^5+(x-5)^5=-32$

L.H.S=(3-x)^5+(x-5)^5

If we put x=5 in L.H.S we get,

(3-5)^5+(5-5)^5=(-2)^5+0=-32=R.H.S