Solving for Sum of Cubes of Reciprocals
This problem involves finding the value of \(\frac{1}{a^3}+\frac{1}{b^3}\) given information about \(\frac{1}{a}+\frac{1}{b}\) and \(\frac{1}{a^2}+\frac{1}{b^2}\). We can simplify this by using substitution and algebraic identities.
Simplifying the Expressions
Let's make the problem easier to handle by introducing new variables:
- Let $x = \frac{1}{a}
- Let $y = \frac{1}{b}
With these substitutions, the given equations become:
- \(x + y = \frac{5}{6}\)
- \(x^2 + y^2 = \frac{13}{36}\)
And the expression we need to find is:
- \(x^3 + y^3\)
Finding the Product \(xy\)
We can use the algebraic identity \((x+y)^2 = x^2 + y^2 + 2xy\) to find the value of \(xy\).
Calculation Steps:
- Start with the identity: \((x+y)^2 = x^2 + y^2 + 2xy\)
- Substitute the given values: \((\frac{5}{6})^2 = \frac{13}{36} + 2xy\)
- Calculate the square: \(\frac{25}{36} = \frac{13}{36} + 2xy\)
- Isolate \(2xy\): \(2xy = \frac{25}{36} - \frac{13}{36}\)
- Perform the subtraction: \(2xy = \frac{12}{36}\)
- Simplify the fraction: \(2xy = \frac{1}{3}\)
- Solve for \(xy\): \(xy = \frac{1}{3} \div 2 = \frac{1}{6}\)
So, we found that \(xy = \frac{1}{6}\).
Calculating the Sum of Cubes \(\frac{1}{a^3}+\frac{1}{b^3}\)
Now, we need to find \(x^3 + y^3\). We can use the identity \(x^3 + y^3 = (x+y)^3 - 3xy(x+y)\).
Calculation Steps:
- Start with the identity: \(x^3 + y^3 = (x+y)^3 - 3xy(x+y)\)
- Substitute the known values (\(x+y = \frac{5}{6}\) and \(xy = \frac{1}{6}\)): \(x^3 + y^3 = (\frac{5}{6})^3 - 3(\frac{1}{6})(\frac{5}{6})\)
- Calculate the terms: \(x^3 + y^3 = \frac{125}{216} - 3(\frac{5}{36})\)
- Simplify the second term: \(x^3 + y^3 = \frac{125}{216} - \frac{15}{36}\)
- To subtract the fractions, find a common denominator, which is 216. Convert \(\frac{15}{36}\): \(\frac{15}{36} = \frac{15 \times 6}{36 \times 6} = \frac{90}{216}\)
- Perform the subtraction: \(x^3 + y^3 = \frac{125}{216} - \frac{90}{216}\)
- Calculate the final result: \(x^3 + y^3 = \frac{125 - 90}{216} = \frac{35}{216}\)
Therefore, the value of \(\frac{1}{a^3}+\frac{1}{b^3}\) is \(\frac{35}{216}\).
Similar Questions
- If \(x^4 = x^2 + 1\), where \(x > 0\), then what is \(2x^4\) equal to?
- If $\frac{\sqrt{p+x}+\sqrt{p-x}}{\sqrt{p+x}-\sqrt{p-x}} = p$, then what is $x$ equal to?
- If \(x-\frac{1}{x}=2\), \(x > 0\); then what is \(x^2-\frac{1}{x^2}\) equal to?
- If \((a-b)^2 + (b - c)^2 + (c-a)^2 = 6\) and \(a^2 + b^2+c^2 = 29\), then what is \((a + b + c)\) equal to?
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- If \(x=2+2^{1/2}+2^{3/2}\), then what is \(x^2-4x-10\) equal to?
Important Questions from Algebric Equations
A group of 630 children is seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible?
The letters L, M, N, 0, P, Q, R, S and T in their order are substituted by nine integers 1 to 9 but not in that order. 4 is assigned to P. The difference between P and T is 5. The difference between N and T is 3.
What is the integer assigned to N?
Four persons, Alok, Bhupesh, Chander and Dinesh have a total of Rs. 100 among themselves. Alok and Bhupesh between them have as much money as Chander and Dinesh between them, but Alok has more money than Bhupesh; and Chander has only half the money that Dinesh has. Alok has in fact Rs. 5 more than Dinesh has.
Who has the maximum amount of money?
If x=3/2, then the value of 27x3-54x2+36x-11 is
If a+b+c = 6 and ab+bc+ca = 11, then the value of bc(b+c) + ca(c+a) +ab(a+b) +3abc is
