Question 86 (x)
Using suitable identities, evaluate the following:
104×97
We have, 104×97=(100+4)(100−3)=(100)2+(4−3)100+4×(−3)=10000+100−12 =10088 [using the identity, (x+a)(x+b)=x2+(a+b)x+ab]
Question 86 (viii)
52×53
Question 86 (iv)
(98)2
Question 86 (xx)
(339)2−(161)2
Question 86 (v)
(1005)2