#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# if a3 + b3 = a-b show that a2 + b2 < 1 if a,b are +ve reals Grade:Upto college level

## 1 Answers AskIITians Expert PRASAD IIT Kharagpur
18 Points
11 years ago

a,b positive reals so

a3 + b3 > 0    ,   so a - b > 0   so,  a > b.

As  a > 0, b > 0,     so, (a+b) > (a - b )   so ,  ((a-b) / (a + b)) < 1.

As,  a3 + b3 = a - b    so, a - a3  =  b + b3    so,  a/b = 1 + b2 / 1 - a2

Sy componendo-dividendo,    we have   (a+b)/(a-b) = (2 - a2 + b2) / (a2 + b2 )

so,  (a+b)/(a-b)  > 1,  we have   (2 - a2 + b2) > (a2 + b2 )which gives,     a2 < 1  so  0 <a <1,  as, b < a, so 0 < b < a < 1

so, ab<1.

a3 + b3  = a - b     given,

so, (a + b)(a2 + b2 - ab) = (a-b)

so,  a2 + b2 - ab  =   (a-b) / (a + b)

from above we have Right Hand Side less than 1,

so, a2 + b2 - ab < 1   so,  a2 + b2 < ab

As, ab < 1,

we have,           a2 + b2 < 1.

Hence Proved.

## ASK QUESTION

Get your questions answered by the expert for free