If sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2 then the value of x100+y100+z100−3x101+y101+z101 is

We have sin–1x + sin–1y + sin–1z = 3π2 it is possible only whensin−1⁡x=π2⇒x=1 ∵sin−1⁡x≤π2sin−1⁡y=π2⇒y=1; sin−1⁡z=π2⇒z=1∴x100+y100+z100−3x101+y101+z101=1+1+1−33=3−1=2