Question

3. If x^2 + y^2 = 29 and x – y = 3, then find the value of x^3– y^3. * 1 point

Answer 1

Gɪᴠᴇɴ :-

• x² + y² = 29
• (x - y) = 3

Tᴏ Fɪɴᴅ :-

• x³ - y³ = ?

Sᴏʟᴜᴛɪᴏɴ :-

→ (x - y) = 3

Squaring Both Sides we get,

→ (x - y)² = 3²

using (a - b)² = a² + b² - 2ab in LHS , we get,

→ x² + y² - 2xy = 9

Putting given value of x² + y² Now,

→ 29 - 2xy = 9

→ 2xy = 29 - 9

→ 2xy = 20

Dividing Both sides by 2,

→ xy = 10.

______________

Now, we Have :-

→ x² + y² = 29

→ x - y = 3

→ xy = 10

we know That, (a³ - b³) = (a - b)(a² + b² + ab)

Therefore,

→ (x³ - y³) = (x - y)(x² + y² + xy)

→ (x³ - y³) = 3(29 + 10)

→ (x³ - y³) = 3 * 39

→ (x³ - y³) = 117 (Ans.)

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star \: {\sf{\green{ {x}^{2} + {y}^{2} = 29}}}$

$\star \: {\sf{\green{ x - y = 3}}}$

${\bf{\blue{\underline{To:Find:}}}}$

$\star \: {\sf{\green{ {x}^{3} - {y}^{3} }}}$

${\sf{\underline{\blue{Now,}}}}$

${\implies{\bf{ x - y = 3}}}$

$\star \: {\bf{ squaring \: both \: side}}$

${\implies{\bf{ (x - y)^{2} = ( {3})^{2} }}}$

${\implies{\bf{ {x}^{2} + {y}^{2} - 2xy = 9}}}$

${\sf{\underline{\blue{we \: \: know \: that \: {x}^{2} + {y}^{2} = 29 ,}}}}$

${\implies{\bf{ 29 - 2xy = 9}}}$

${\implies{\bf{ - 2xy = 9 - 29}}}$

${\implies{\bf{ - 2xy = - 20}}}$

${\implies{\bf{ xy = \frac{20}{2} }}}$

${\implies{\bf{ \boxed{ xy = 10}}}}$

${\sf{\underline{\blue{Now,}}}}$

${\star \: {\bf{ {x}^{3} - {y}^{3} = ( x - y)( {x}^{2} + {y}^{2} + xy)}}}$

${\implies{\bf{ (x - y) \big[( {x}^{2} + {y}^{2}) + (xy) \big]}}}$

${\implies{\bf{ (3) \big[( 29) + (10)]}}}$

${\implies{\bf{ (3) \times \big[39]}}}$

${\implies{\bf{ 117}}}$

${\bf{\blue{\underline{Hence \: \: \: {x}^{3} - {y}^{3} = 117 Ans}}}}$