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If x + y + z = 19, xy + yz + zx = 144, then the value of $$\sqrt {{x^3} + {y^3} + {z^3} - 3xyz} $$ is:
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& x + y + z = 19, \cr
& xy + yz + zx = 144, \cr
& \sqrt {{x^3} + {y^3} + {z^3} - 3xyz} \cr
& {\text{Let }}z = 0 \cr
& x + y = 19,\,xy = 144,\,\sqrt {{x^3} + {y^3}} = ? \cr
& \sqrt {{x^3} + {y^3}} \cr
& = \sqrt {\left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} - 3xy} \right]} \cr
& = \sqrt {19\left( {{{19}^2} - 3 \times 144} \right)} \cr
& = \sqrt {19 \times 19} \cr
& = 19 \cr} $$
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