Let p(x)=x3+3x2+3x+1 (i) The zero of x + 1 is –1.∴p(−1)=(−1)3+3(−1)2+3(−1)+1 =–1 + 3 – 3 + 1 = 0Thus, the required remainder = 0(ii) The zero of x−21​ is 21​ ∴p(21​) =(21​)3+3(21​)2+3(21​)+1 =81​+43​+23​+1=81+6+12+8​=827​ Thus, the required remainder =827​ (iii) The zero of x is 0 .∴p(0)=(0)3+3(0)2+3(0)+1 = 0 + 0 + 0 + 1 =1Thus, the required remainder =1.(iv) The zero of x+π is −π.p(−π)=(−π)3+3(−π)22+3(−π)+1 =−π3+3π2+(−3π)+1 =−π3+3π2−3π+1 Thus, the required remainder is −π3+3π2−3π+1 (v) The zero of 5 + 2x is −25​.∴p(−25​)=(−25​)3+3(−25​)2+3(2−5​)+1 =−8125​+3(425​)+(−215​)+1 =8−125​+475​−215​+1=8−27​ Thus, the required remainder is −827​.