if x = cube root 28 and y = cube root 27, Find the value of x + y - (1 / x^2 + xy + y^2). please solve it if you know
Answers
\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=6}x+y−x2+xy+y21=6
Step-by-step explanation:
Given
\mathbf{x=\sqrt[3]{28}}x=328
So \mathbf{x^{3}=28}x3=28 ...1)
\mathbf{y=\sqrt[3]{27}=3}y=327=3
So \mathbf{y^{3}=27}y3=27 ...2)
First consider
\mathbf{\frac{1}{x^{2}+xy+y^{2}}}x2+xy+y21 ...3)
Multiply by (x-y) in numerator and denominator, we get
\mathbf{\frac{x-y}{(x-y)(x^{2}+xy+y^{2})}}(x−y)(x2+xy+y2)x−y
Denominator of above term can be written as
\mathbf{\frac{x-y}{x^{3}-y^{3}}}x3−y3x−y ...4)
From equation 1), equation 2) and equation 4)
\mathbf{\frac{x-y}{28-27}=x-y}28−27x−y=x−y ...5)
Means from equation 3) and equation 5)
\mathbf{\frac{1}{x^{2}+xy+y^{2}}=x-y}x2+xy+y21=x−y ...6)
Now come to question
\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}}x+y−x2+xy+y21 ...7)
From equation 6) , equation 7) can be written as
\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=x+y-(x-y)}x+y−x2+xy+y21=x+y−(x−y)
\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=x+y-x+y}x+y−x2+xy+y21=x+y−x+y
\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=2y}x+y−x2+xy+y21=2y (where y=3)
\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=6}x+y−x2+xy+y21=6
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