Question

Evalute the following using suitable identity (i) (99)^3, (ii) (102)^3, (iii) (998)^3

Answer 1

these sums will be solved using the identity

(a+b) ^3 = a^3 + b^3 + 3ab (a + b)

for eg., in 1st one =
(100 - 1)
here u will have to use (a - b)^3 = ^3 - b^3 - 3ab ( a-b )

hope it helps
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Answer 2

$\mathbb{ANSWER}$ :

(i) (99)³

$\boxed{\underline { Method.1}}$

(100 - 1)³

[ $\therefore$ using (x-y)³ = x³ - y³ - 3xy(x-y) ]

= (100)³ - (1)³ - 3*100*1 (100 - 1)

= 1000000 - 1 - 300*99

= 1000000 - 29701

= 970299

.

$\boxed{\underline { Method.2}}$

(90+9)³

[ $\therefore$ using (x+y)³ = x³ + y³ + 3xy(x+y) ]

= (90)³ + (9)³ + 3*90*9 (90+9)

= 729000 + 729 + 2430*99

= 729000 + 729 + 240570

= 970299


➖➖➖➖➖➖➖➖➖➖➖

(ii) (102)³

$\boxed{\underline { Method.1}}$

(100+2)³

[ $\therefore$ using (x+y)³ = x³ + y³ + 3xy(x+y) ]

= (100)³ + (2)³ + 3*100*2 (100+2)

= 1000000 + 8 + 600*102

= 1000000 + 8 + 61200

= 1061208

.

$\boxed{\underline { Method.2}}$

(110-8)³

[ $\therefore$ using (x-y)³ = x³ - y³ - 3xy(x-y) ]

= (110)³ - (8)³ - 3*110*8 (110-8)

= 1331000 - 512 - 2640*102

= 1331000 - 512 - 269280

= 1331000 - 269792

= 1061208

➖➖➖➖➖➖➖➖➖➖➖


(iii) (998)³

$\boxed{\underline { Method.1}}$

(1000-2)³

[ $\therefore$ using (x-y)³ = x³ - y³ - 3xy(x-y) ]

= (1000)³ - (2)³ - 3*1000*2 (1000-2)

= 1000000000 - 8 - 6000*998

= 1000000000 - 8 - 5988000

= 1000000000 - 5988008

= 99401192

.

$\boxed{\underline { Method.2}}$

(990+8)³

[ $\therefore$ using (x+y)³ = x³ + y³ + 3xy(x+y) ]

= (990)³ + (8)³ + 3*990*8 (990+8)

= 970299000 + 512 + 23760*998

= 970299000 + 512 + 23712480

= 99401192