Given,
⇒x=3m+1−3m−13m+1+3m−1
Applying componendo and dividendo, we get :
⇒x−1x+1=3m+1+3m−1−(3m+1−3m−1)3m+1+3m−1+3m+1−3m−1⇒x−1x+1=3m+1+3m−1−3m+1+3m−13m+1+3m−1+3m+1−3m−1⇒x−1x+1=23m−123m+1⇒x−1x+1=3m−13m+1
Cubing both sides, we get :
⇒(x−1x+1)3=m−1m+1⇒(x+1)3(m−1)=(x−1)3(m+1)⇒(x3+3x2+3x+1)(m−1)=(x3−3x2+3x−1)(m+1)⇒mx3+3mx2+3mx+m−x3−3x2−3x−1=mx3−3mx2+3mx−m+x3−3x2+3x−1⇒mx3+3mx2+3mx+m−x3−3x2−3x−1−(mx3−3mx2+3mx−m+x3−3x2+3x−1)=0⇒mx3−mx3+3mx2+3mx2+3mx−3mx+m+m−x3−x3−3x2+3x2−3x−3x−1+1=0⇒6mx2+2m−2x3−6x=0⇒2(3mx2+m−x3−3x)=0⇒3mx2+m−x3−3x=0⇒x3−3mx2+3x−m=0.
Hence, proved that x3 - 3x2m + 3x - m = 0.