Question

THE VALUE OF (262) CUBE+(238)CUBE ÷(262)^-262×238+(238)^

a)400
b)500
c)300
d) none of the above ​

Answer 1

To find -

$\sf{ \bold { Find \: the \: value \: of \: - }} \\ \\ \sf{ \implies { \dfrac{ ( 262 )^3 + ( 238 )^3 }{ ( 262 )^2 - 262 \times 238 + ( 238 )^2 } } } \\ \\ \sf{ \bold { \star Solution \: - }} \\ \\ \sf{ \implies { \dfrac{ ( 262 )^3 + ( 238 )^3 }{ ( 262 )^2 - 262 \times 238 + ( 238 )^2 } } } \\ \\ \sf{ \implies { \dfrac{ ( 262 + 238 )( (262 )^2 - 262 \times 238 + ( 238 )^2 ) }{ ( 262 )^2 - 262 \times 238 + ( 238 )^2 } }} \\ \\ \sf{ \implies { \dfrac{ ( 262 + 238 ) \not { \cancel{ ( (262 )^2 - 262 \times 238 + ( 238 )^2 ) } } }{ \not { \cancel{ ( 262 )^2 - 262 \times 238 + ( 238 )^2 } } } }} \\ \\ \sf{ \implies { 238 + 262 }} \\ \\ \sf{ \implies { 500 }}$ .

Hence , Option B is the correct answer.

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Additional Information -

( a + b )² = a² + 2ab + b²

( a - b )² = a² - 2ab + b²

( a + b )( a - b ) = a² - b²

( a + b )³ = a³ + 3ab ( a + b ) + b³

( a - b )³ = a³ - 3ab ( a + b ) - b³

( a + b + c )³ = a³ + b³ + c³ + 3 ( a + b )( b + c )( c + a )

a³ + b³ + c³ - 3abc = ( a + b + c )( a² + b² + c² - ab - bc - ca )

When a + b + c = 0 ,

a³ + b³ + c³ = 3abc .

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