If x = \sqrt[3]{28} and y = \sqrt[3]{27}, find the value of x + y - \frac{1}{x^2 + xy + y^2}
Answer (2)
To calculate x + y − x 2 + x y + y 2 1 with x = 3 28 and y = 3 27 , first find x + y , then compute x 2 + x y + y 2 , and finally substitute back into the expression for evaluation. The process requires approximation for accurate results. Ultimately, numeric or calculator results will facilitate reaching an answer.
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To solve the problem, we need to find the value of the expression x + y − x 2 + x y + y 2 1 given that x = 3 28 and y = 3 27 .
Step-by-Step Solution:
Evaluate x and y :
x = 3 28 is the cube root of 28.
y = 3 27 is the cube root of 27. Since 27 is a perfect cube, we have y = 3 .
Calculate x + y :
Using the approximate value for x , which is slightly more than 3 (because 27"> 28 > 27 ), we can say x ≈ 3.036 .
Therefore, x + y ≈ 3.036 + 3 = 6.036 .
Calculate the denominator x 2 + x y + y 2 :
x 2 ≈ ( 3.036 ) 2 ≈ 9.217 .
x y ≈ 3.036 × 3 = 9.108 .
y 2 = 3 2 = 9 .
So, x 2 + x y + y 2 ≈ 9.217 + 9.108 + 9 = 27.325 .
Calculate the fraction x 2 + x y + y 2 1 :
27.325 1 ≈ 0.0366 .
Final Expression x + y − x 2 + x y + y 2 1 :
Substitute these into the expression: 6.036 − 0.0366 ≈ 5.9994 .
Approximating further, this is very close to 6.
Thus, x + y − x 2 + x y + y 2 1 ≈ 6 . This slight approximation can be due to rounding and using approximate values for non-perfect cubes.
