If(3y−3y)=5,\left ( 3y-\frac{3}{y} \right )=5,(3y−y3)=5, find the value of (y2+1y2)\left ( y^2+\frac{1}{y^2} \right )(y2+y21). A479\frac{47}{9}947 B499\frac{49}{9}949 C419\frac{41}{9}941 D439\frac{43}{9}943 Show answer & explanationCorrect answer: DExplanationThe correct answer is Option 4 i.e. 439\frac{43}{9}943. ⇒(3y−3y)=5,\left ( 3y-\frac{3}{y} \right )=5,(3y−y3)=5, ⇒ 3(y−1y)=53( y-\frac{1}{y} ) = 53(y−y1)=5 ⇒ (y−1y)=53( y-\frac{1}{y} ) = {5\over3}(y−y1)=35 Squaring both sides ⇒ (y−1y)2=(53)2( y-\frac{1}{y})^2 = ({5\over3})^2(y−y1)2=(35)2 ⇒ (y2+1y2−2×y×1y)=259{( y^2 + {1\over y^2} - {2\times y\times{1\over y}}}) = {25\over9}(y2+y21−2×y×y1)=925 ⇒ (y2+1y2−2)=259{( y^2 + {1\over y^2} - {2}}) = {25\over9}(y2+y21−2)=925 ⇒ (y2+1y2)=259+2{( y^2 + {1\over y^2} }) = {25\over9} +2(y2+y21)=925+2 ⇒ (y2+1y2)=25+189{( y^2 + {1\over y^2} }) = {{25 + 18}\over9 }(y2+y21)=925+18 ⇒ (y2+1y2)=439{( y^2 + {1\over y^2} }) = {{43}\over9 }(y2+y21)=943