If 1 + 9r² + 81 r² = 256 and 1 + 3r+9r² = 32, then find the value of 1- 3r+9r².
Answers
Given : 1 + 9r² + 81 r⁴ = 256 and 1 + 3r+9r² = 32,
To Find : the value of 1- 3r+9r².
4
8
16
12
Solution:
Assume that
1- 3r+9r². = k
1 + 3r+9r² = 32
multiply both
=> ((1 + 9r² ) - 3r)((1 + 9r² ) + 3r) = 32k
=> (1 + 9r²)² - (3r)² = 32k
=> 1 + 81r⁴ + 18r² - 9r² = 32k
=> 1 + 81r⁴ + 9r² = 32k
=> 256 = 32k
=> k = 8
=> 1 - 3r+9r². = 8
Additional Info : Data is inconsistent :
1 + 3r+9r² = 32
1 - 3r+9r². = 8
on subtracting 6r = 24 => r = 4
on adding
1 + 9r² = 20
=> 9r² = 19 where r ≠ 4
Another method :
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
a = 1
b = 3r
c = 9r²
(1 + 3r + 9r²)² = 1 + 9r² + 81 r⁴ + 2 (3r + 27r³ + 9r²)
=> 32² = 256 + 2 (3r + 27r³ + 9r²)
=> 384 = (3r + 27r³ + 9r²)
=> 128 = r(1 + 9r² + 3r)
=> 128 = r ( 32)
=> r = 4
but r = 4 does not satisfy any of the equation
any ways 1 - 3r+9r². = 8
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