(9)/(b)+(b)/(a)=-1" them "a^(3)+b^(3)

If a+b+c=1,a^(2)+b^(2)+c^(2)=9 and a^(3)+b^(3)+c^(3)=1, then (1)/(a)+(1)/(b)+(1)/(c) is (i)0 (ii) -1(iii)1(iv)3

sin ^ (- 1) (a- (a ^ (2)) / (3) + (a ^ (3)) / (9) + (a ^ (3)) / (9) +) + cos ^ ( -1) (1 + b + b ^ (2) +) = (pi) / (2) sin a - + ... + cos11 + b + b2 + ...

find(a+b)-:(a-b) if (i)a=(1)/(3),b=(3)/(5)(i)a=-(1)/(13),b=(4)/(9)(iii)a=-(6)/(7)

The following steps are involved in finding the value of 10 (1)/(3) xx 9(2)/(3) by using an appropriate indentity . Arrange them in sequential order . (A) (10)^(2) - ((1)/(3))^(2) = 100 - (1)/(9) (B) 10(1)/(3) xx 9(2)/(3) = (10 + (1)/(3)) (10 - (1)/(3)) (C) (10 + (1)/(3)) (10 - (1)/(3)) = (10)^(2) - ((1)/(3))^(2) [because (a + b) (a -b) = (a^(2) - b^(2))] (D) 100 - (1)/(9) = 99 + 1 - (1)/(9) = 99(8)/(9)

If(1,a),(3,9a),(4,b),(6,18) then (a,(b)/(a))=

If lim_(x rarr0)(x^(-3)sin3x+ax^(-2)+b) exists and is equal to 0, then (a)a=-3 and b=(9)/(2)(b)a=3 and b=(9)/(2)(c)a=-3 and b=-(9)/(2) (d) a=3 and b=-(9)/(2)

Simplify each of the products: ((1)/(2)a-3b)(3b+(1)/(2)a)((1)/(4)a^(2)+9b^(2))