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Question
Show that the value of x is 11, when `("x"^3 + "3x")/(3"x"^2 + 1) = 341/91`
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Solution
`("x"^3 + "3x")/(3"x"^2 + 1) = 341/91`
⇒ 91(x3 + 3x) = 341(3x2 + 1)
⇒ 91x3 - 1023x2 + 273x - 341 = O
LHS
91x3 - 1023x2 + 273x - 341
= 91( 11 × 11 × 11) - 1023 (11 × 11) + 273 (11) -341
= (91 × 1331) - (1023 × 121) + (273 × 11) - 341
= 121121 - 123783 + 3003 - 341
= 0= RHS
LHS = RHS.
Hence x = 11
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