C-03-Q142mediumsingle_mcqFactorize f(a)=a3−9+(a+1)3f(a) = a^3 - 9 + (a+1)^3f(a)=a3−9+(a+1)3:a(a−1)(2a2+5a+8)(a-1)(2a^2 + 5a + 8)(a−1)(2a2+5a+8)b(a+1)(2a2−5a+8)(a+1)(2a^2 - 5a + 8)(a+1)(2a2−5a+8)c(a−1)(a2+5a−8)(a-1)(a^2 + 5a - 8)(a−1)(a2+5a−8)d(a−1)3(a-1)^3(a−1)3ব্যাখ্যাExpanding gives a3−9+a3+3a2+3a+1=2a3+3a2+3a−8a^3 - 9 + a^3 + 3a^2 + 3a + 1 = 2a^3 + 3a^2 + 3a - 8a3−9+a3+3a2+3a+1=2a3+3a2+3a−8. Since a=1a=1a=1 makes this zero, (a−1)(a-1)(a−1) is a factor, and dividing leaves 2a2+5a+82a^2 + 5a + 82a2+5a+8.