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805+ (Hard)|   Algebra|      
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Bunuel
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 21 Sep 2020, 02:45
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C
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95% (hard)

Question Stats:

48% (02:51) correct 52% (02:47) wrong based on 118 sessions

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If \(x = 1 + \sqrt[3]{5} + \sqrt[3]{5^2}\), then the value of \(x^3 - 3x^2 - 12x + 6 \) is

A. 20
B. 21
C. 22
D. 23
E. 25



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C
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 21 Sep 2020, 03:13
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Using Approximation
X = 1+5^1/3+5^2/3
X = 1+1.7+2.9
X = 5.63

Therefore, X^3-3x^2-12x+6 =
=> (5.63)^3 - 3(5.63)^2 - 12(5.63) + 6
=> 178.45 - 3(31.69) - 12(5.63) + 6
=> 178 - 95 - 67 + 6
=> 22

Answer is C

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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 21 Sep 2020, 03:24
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Explanation:
x=1+ 3√5 + 3√25
x-1 = (5)^1/3 + (5)^2/3 --- Eqn 1
Cube above eqn 1
(x-1)^3 = {(5)^1/3 + (5)^2/3}^3
x^3 - 1 -3*(x)*(x-1) = (5) + (5)^2 + 3*{(5)^1/3}*{(5)^2/3}*{(5)^1/3 + (5)^2/3}
x^3 - 1 - 3x^2 + 3x = 5 + 25 +15*{(5)^1/3 + (5)^2/3}
From Eqn 1, x^3 - 3x^2 + 3x - 1 = 30 + 15(x-1)
x^3 - 3x^2 -12x - 1 + 7 = 15 + 7
x^3 − 3x^2 − 12x + 6 = 22

IMO-C
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 21 Sep 2020, 04:17
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\((a-b)^3 = a^3 - b^3 - 3ab(a-b)\)
\((a+b)^3 = a^3 + b^3 + 3ab(a+b)\)

Bunuel
If \(x = 1 + \sqrt[3]{5} + \sqrt[3]{5^2}\), then the value of \(x^3 - 3x^2 - 12x + 6 \) is

A. 20
B. 21
C. 22
D. 23
E. 25
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\(x = 1 + \sqrt[3]{5} + \sqrt[3]{5^2}\)
\(x-1 = 5^⅓ + 5^⅔\)
\((x-1)^3 = (5^⅓ + 5^⅔)^3\)

Applying the blue identity above to the left side and the red identity above to the right side, we get:
\(x^3 - 1^3 - 3*x*1*(x-1) = (5^⅓)^3 + (5^⅔)^3 + 3*5^⅓ * 5^⅔ * (5^⅓ + 5^⅔)\)
\(x^3 - 1 - 3x^2 + 3x = 5 + 25 + 15(5^⅓ + 5^⅔)\)
\(x^3 - 3x^2 + 3x - 1 = 30 + 15(5^⅓ + 5^⅔)\)

As noted in the green equation above, \(x-1 = 5^⅓ + 5^⅔\).
Replacing \(5^⅓ + 5^⅔\) with \(x-1\), we get:
\(x^3 - 3x^2 + 3x - 1 = 30 + 15(x-1)\)
\(x^3 - 3x^2 + 3x - 1 = 30 + 15x - 15\)
\(x^3 - 3x^2 - 12x - 1 = 15\)

Adding 7 to both sides, we get:
\(x^3 - 3x^2 - 12x + 6 = 22\)

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C

This problem is beyond the scope of the GMAT.
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 21 Sep 2020, 05:50
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If \(x = 1 + \sqrt[3]{5} + \sqrt[3]{5^2}\), then the value of \(x^3 - 3x^2 - 12x + 6 \) is

A. 20
B. 21
C. 22 --> correct
D. 23
E. 25

Solution:
\(x = 1 + \sqrt[3]{5} + \sqrt[3]{5^2}\)
=> \(x - 1 = \sqrt[3]{5} + \sqrt[3]{5^2}\)
=> \((x - 1)^3 = (\sqrt[3]{5} + \sqrt[3]{5^2})^3\)
=> \( x^3 -3x^2+3x-1= 5+3 \sqrt[3]{5^2} *\sqrt[3]{5^2} + 3 \sqrt[3]{5}*\sqrt[3]{5^4}+5^2\)
=> \( x^3 -3x^2 = 31+3 \sqrt[3]{5^4} + 3\sqrt[3]{5^5} -3x= 31+15 \sqrt[3]{5} + 15\sqrt[3]{5^2} -3x \) ---(i)
replacing the value of \( x^3 -3x^2 \) from (i) in \(x^3 - 3x^2 - 12x + 6 \)
=\( 31+15 \sqrt[3]{5} + 15\sqrt[3]{5^2} -3x - 12x + 6 \)
=\( 31+15 \sqrt[3]{5} + 15\sqrt[3]{5^2} - 15x + 6 \)
=\( 31+15 \sqrt[3]{5} + 15\sqrt[3]{5^2} -15(1 + \sqrt[3]{5} + \sqrt[3]{5^2}) + 6 \) , replacing the value of x from question
=\( 37+15 \sqrt[3]{5} + 15\sqrt[3]{5^2} -15 -15\sqrt[3]{5} -15\sqrt[3]{5^2} \)
=22
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 14 Jul 2024, 04:50
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(x-1)=5^1/3+5^2/3
=>(x-1)^3=5+5^2+3.5(x-1)

=> x^2-3x^2+3x-1=15x+15

Expr=16+6=22
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6 06 Nov 2025, 02:12
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