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We can also take the approach as follows:

1. For f(x) = 1, we need the expression in the bracket to be equal to 1/A.
2. (x-B)^3, looking at the expression we need +B to negate B and rest in as the cube root of 1/A.
3. Hence the answer is B+1/(A)^1/3 (Option E)

This solution might be a different approach but saves :)
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f(x) = A (x - B)^3 ; A*B not equal to "0"
x= ? when f(x) = 1

f(x) = A(x- B)^3 = 1
=> (x-B)^3 = 1/ A
=> x-B = 1 / (A)^1/3
=> x = 1/ (A)^1/3 + B


Answer : E
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If f(x)=A(x−B)^3, where AB≠0, what is value of x for which f(x)=1?


A. B−A√3B−A3

B. B−1A√3B−1A3

C. B+A√3B+A3

D. −B+1A√3−B+1A3

E. B+1A√3

SOL: given: a, b != 0 and f(x)=1

f(x) = A(x-B)^3 = 1
=> (x-B)^3 = 1/A (taking cube root on both sides)
=> (x-B) = (1/A)^1/3
=> x = B + (1/A)^1/3

Ans is E
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Bunuel
If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?


A. \(B - \sqrt[3]{A}\)

B. \(B - \frac{1}{\sqrt[3]{A}}\)

C. \(B + \sqrt[3]{A}\)

D. \(-B + \frac{1}{\sqrt[3]{A}}\)

E. \(B + \frac{1}{\sqrt[3]{A}}\)

Solution:

A(x - B)^3 = 1

(x - B)^3 = 1/A

x - B = 3^√(1/A)

x = B + 3^√(1/A) = B + 1/(3^√A)

Answer: E
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Bunuel
If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?


A. \(B - \sqrt[3]{A}\)

B. \(B - \frac{1}{\sqrt[3]{A}}\)

C. \(B + \sqrt[3]{A}\)

D. \(-B + \frac{1}{\sqrt[3]{A}}\)

E. \(B + \frac{1}{\sqrt[3]{A}}\)

Asked: If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?

\(f(x) = A(x-B)^3 = 1\)
\((x-B)^3 = 1/A\)
\(x = B + \frac{1}{\sqrt[3]{A}\)

IMO E
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\(f(x) = A(x−B)^3=1\)
\((x−B)^3=\frac{1}{A}\)
\(x−B=\frac{1}{\sqrt[3]{A}}\)
\(x=B-\frac{1}{\sqrt[3]{A}}\)

Answer E
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