Difference between revisions of "1963 AHSME Problems/Problem 40"
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+ | ==Solution 4== | ||
+ | We consider the formula <math>(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3</math>. Factoring out <math>3ab</math>, and the commutative property gives <math>( | ||
+ | a^3 - b^3) - 3ab(a-b)</math>. Cubing both sides gives us | ||
+ | <cmath>(\sqrt[3]{x+9}-\sqrt[3]{x-9})^3=27</cmath> | ||
+ | Using our formula, we have | ||
+ | <cmath>18 - 3 \sqrt[3]{x^2 - 81} = 27</cmath> | ||
+ | Solving this gives <math>x^2 = 80</math>, therefore, the answer is <math>\boxed{\textbf{(C)}}</math>. | ||
+ | ~zixuan12 | ||
==See Also== | ==See Also== |
Latest revision as of 15:45, 17 January 2021
Problem
If is a number satisfying the equation , then is between:
Solution 1
Let and . Cubing these equations, we get and , so . The left-hand side factors as
However, from the given equation , we get . Then , so .
Squaring the equation , we get . Subtracting this equation from the equation , we get , so . But and , so , so . Cubing both sides, we get , so . The answer is .
Solution 2
i.e,
if the sum of three numbers is zero, then their sum of cubes is thrice the product of each number. then, . by solving this, we get . this gives the step what we had done in solution 1.The answer is .
Solution 3
Cubing both sides, we get so Dividing both sides by 3 and cubing, we find , which is between .
~pfalcon
Solution 4
We consider the formula . Factoring out , and the commutative property gives . Cubing both sides gives us Using our formula, we have Solving this gives , therefore, the answer is . ~zixuan12
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Last Problem | |
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