If (9root3 + 11root2)^1/3 = roota +rootb then, A a=root3 B a=3 C b=3 D b=root2 Share with your friends Share 0 Akash Kumar answered this We have 93+11213=a+b⇒ 93+112=a+b3⇒aa+bb+3aba+b= 93+112⇒aa+bb+3ab+3ba= 93+112⇒a+3ba+b+3ab= 93+112⇒a+3ba=93 and b+3ab=112 Or a+3ba=112 and b+3ab=93 ⇒a=3, b=2 Or a=2 , b=3 Hence a=3 or b=3 are possible values. so either of (B) or (C) are correct option 1 View Full Answer
