Click to Chat

1800-2000-838

+91-120-4616500

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
```        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

SHAIK AASIF AHAMED
74 Points
```										Hello student,Please find the answer to your question belowwe 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^3-3abc8x^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 = +1hence 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 ?

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details