Math, asked by abcd8995, 1 year ago

plz answer the question........

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Answered by Anonymous
1
As a,b,c,d are odd perfect cube
then cube root of a,b,c,d will also be a odd number
hence
 \sqrt[3]{a}    \: and \:  \sqrt[3]{b}  \: are \: odd \\  \\    \sqrt[3]{a}  +   \sqrt[3]{b}  \: is \:even \:  \\ as \: sum \: of \: two \: odd \: number \: is \: always \: even \\ hence \: it \: will \: have \: 2 \: as \: a \: factor \\ now \: squaring \: it \: will \: give \: 4 \: as \: a \: factor \\ similarly \\   \sqrt[3]{c}  +  \sqrt[3]{d} is \: even \\ hence \: it \: will \: also \: has \: 2 \: as \: a \: factor \\ multiplying \\ you \: get \\ 4 \times 2 = 8 \: as \: a \: factor
hence 8 will always be the factor of given expression
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