-
If 
x + 1 
2 = 3 , then the value of ( x206 + x200 + x90 + x84 + x18 + x12 + x6 + 1 ) is x
-
- 0
- 1
- 84
- 206
Correct Option: A
Using Rule 8,
| x + | | 2 | = 3 | ||
| x |
| ⇒ x + | = √3 | ||
| x |
On cubing both sides,
| ⇒ | | x + | | 3 | = 3√3 | |
| x |
| ∴ x3 + | + 3 | | x + | | = 3√3 | ||
| x3 | x |
| ⇒ x3 + | + 3√3 - 3√3 = 0 | |
| x3 |
| ⇒ x3 + | = 0 ⇒ x6 + 1 = 0 | |
| x3 |
Now , x206 + x200 + x90 + x84 + x18 + x12 + x6 + 1 = x200( x6 + 1 ) + x84( x6 + 1 ) + x12( x6 + 1 ) + ( x6 + 1 )
= 0
