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
Menu
ajinkya bhole Grade: 10
        Factorize: 8x^6 + 5x^3 + 1(use a^3 + b^3 + c^3 -3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac)

(^-raised to)
7 years ago

Answers : (1)

SHAIK AASIF AHAMED
askIITians Faculty
74 Points
										Hello student,
Please find the answer to your question below
we checking rational solutions find that there is no rational root and hence there is no term of the form (x+a)
now 8x^6 = (2x^2)^3 and 1= 1^3 if we can express 5x^3 so that one is a cube and second one is product of -x 2 and the 3rd term then we get into the form a^3+b^3+c^3-3abc
8x^6+5x^3+1 = 8x^6+(-x)^3 + 1^3 +6x^3
= 8x^6 - x^3 + 1^3 + 3(x^2)(x)
= a^3+b^3+c^3 - 3abc where a = 2x^2 , b = -x and c = +1
hence it factors as (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (2x^2 -x +1)(4x^4+x^2+1 + 2x^3 +2x -2x^2)
=(2x^2-x+1)(4x^4+2x^3-x^2+2x+1)
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details