Question

y= 7 + 4 √(3) x= 7 - 4 √(3)
find x cube + y cube

Answer 1

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Given,


y = 7 + 4√3

x = 7 - 4√3.

Now,

= x³ + y³

= ( x + y ) ( x² + y² - xy ).

We have the values of x and y but not the values of x² , y² and xy.So, first we need to find them.

⇒ x² = ( 7 + 4√3 )²

⇒ x² = ( 7 )² + ( 4√3 )² + 2 × 7 × 4√3

⇒ x² = 49 + 48 + 56√3

∴ x² = 97 + 56√3

Now,

⇒ y² = ( 7 - 4√3 )²

⇒ y² = ( 7 )² + ( 4√3 )² - 2 × 7 × 4√3

⇒ y² = 49 + 48 - 56√3

∴  y² = 97 - 56√3

Now,

⇒ xy = ( 7 + 4√3 ) ( 7 - 4√3 )

⇒ xy = ( 7 )² - ( 4√3 )²

⇒ xy = 49 - 48

∴  xy = 1.

Now,

= ( x + y ) ( x² + y² - xy )

= ( 7 + 4√3 + 7 - 4√3 ) ( 97 + 56√3 + 97 - 56√3 - 1 )

= ( 14 ) ( 194 - 1 )

= 14 × 193

= 2,702.



Hope it helps !

Answer 2

heya friend here's ur answer :-)

$y = 7 + 4 \sqrt{3} \\ \\ \frac{1}{y} = \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } = \frac{ 7 - 4\sqrt{3} }{(7) {}^{2} - (4 \sqrt{3}) {}^{2} } = \frac{7 - 4 \sqrt{3} }{49 - 48} = 7 - 4 \sqrt{3} \\ \\ x = 7 - 4 \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } = \frac{7 + 4 \sqrt{3} }{(7) {}^{2} - (4 \sqrt{3}) {}^{2} } = 7 + 4 \sqrt{3}$


now,

$x + y = (7 - 4 \sqrt{3} ) + (7 + 4 \sqrt{3} ) \\ x + y = 14 \\ \\ (x + y) {}^{3} = x {}^{3} + y {}^{3} + 3xy(x + y) \\ (14) {}^{3} = x {}^{3} + y {}^{3} + 3 \times x \times y(14) \\ x {}^{3} + y {}^{3} = 51xy - 2744 \\ x {}^{3} + y {}^{3} = 51{(7 - 4 \sqrt{3} ) (7 + 4 \sqrt{3} )} - 2744\\ x {}^{3} + y {}^{3} = 357 - 204 \sqrt{3 } + 357 + 204 \sqrt{3 } - 2744 \\ x {}^{3} + y {}^{3} = 714 - 2744 \\ x {}^{3} + y {}^{3} = - 2030$

hope it helps