Question

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Please answer my Questionn

if x = $\sf{\dfrac{1}{7 + 4\sqrt{3}}$ , y = $\sf{\dfrac{1}{7 - 4\sqrt{3}}$ then find the Value of x³ + y³

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Answer 1

Question :-

If $\bf{x = \dfrac{1}{7 + 4\sqrt{3}}}$ and $\bf{y = \dfrac{1}{7 - 4\sqrt{3}}}$ then find the Value of $\bf{x^{3} + y^{3}}$ .

To Find :-

The value of $\bf{x^{3} + y^{3}}$

Given :-

• $\bf{x = \dfrac{1}{7 + 4\sqrt{3}}}$

• $\bf{y = \dfrac{1}{7 - 4\sqrt{3}}}$

We Know :-

$\boxed{\bf{a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})}}$

Solution :-

By using the formula and substituting the values in it , we get :-

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

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)}$$\bf{\bigg[\bigg(\dfrac{1}{7 + 4\sqrt{3}}\bigg)^{2} - \bigg(\dfrac{1}{7 + 4\sqrt{3}} \times \dfrac{1}{7 - 4\sqrt{3}}\bigg) + \bigg(\dfrac{1}{7 - 4\sqrt{3}}\bigg)^{2}\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{1}{7 + 4\sqrt{3}}\bigg)^{2} - \bigg(\dfrac{1}{7^{2} - (4\sqrt{3})^{2}}\bigg) + \bigg(\dfrac{1}{7 - 4\sqrt{3}}\bigg)^{2}\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{1}{7 + 4\sqrt{3}}\bigg)^{2} - \bigg(\dfrac{1}{49 - 48}\bigg) + \bigg(\dfrac{1}{7 - 4\sqrt{3}}\bigg)^{2}\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{1}{7 + 4\sqrt{3}}\bigg)^{2} - \bigg(\dfrac{1}{1}\bigg) + \bigg(\dfrac{1}{7 - 4\sqrt{3}}\bigg)^{2}\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{1}{7 + 4\sqrt{3}}\bigg)^{2} - 1 + \bigg(\dfrac{1}{7 - 4\sqrt{3}}\bigg)^{2}\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)}$$\bf{\bigg[\bigg(\dfrac{1}{7^{2} + (2 \times 7 \times 4\sqrt{3}) + (4\sqrt{3})^{2}}\bigg) - 1 + \bigg(\dfrac{1}{7^{2} - (2 \times 7 \times 4\sqrt{3}) + (4\sqrt{3})^{2}}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{1}{49 + 56\sqrt{3} + 48}\bigg) - 1 + \bigg(\dfrac{1}{49 - 56\sqrt{3} + 48}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{1}{97 + 56\sqrt{3}}\bigg) - 1 + \bigg(\dfrac{1}{97 - 56\sqrt{3}}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{97 - 56\sqrt{3} - [97^{2} - (56\sqrt{3})]^{2} + 97 + 56\sqrt{3}}{97^{2} - (56\sqrt{3})^{2}}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{97 - [97^{2} - (56\sqrt{3})^{2}] + 97}{97^{2} - (56\sqrt{3})^{2}}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{194 - [97^{2} - (56\sqrt{3})^{2}]}{97^{2} - (56\sqrt{3})^{2}}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{194 - [9409 - 9408]}{9409 - 9408}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg)\bigg[\bigg(\dfrac{194 - 1}{1}\bigg)\bigg]}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{1}{7 + 4\sqrt{3}} + \dfrac{1}{7 - 4\sqrt{3}}\bigg) \times 193}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{7 - 4\sqrt{3} + 7 + 4\sqrt{3}}{7^{2} - (4\sqrt{3})^{2}}\bigg) \times 193}$

$:\implies \bf{a^{3} + b^{3} = \bigg(\dfrac{7 - 4\sqrt{3} + 7 + 4\sqrt{3}}{49 - 48}\bigg) \times 193}$

$:\implies \bf{a^{3} + b^{3} = (7 - 4\sqrt{3} + 7 + 4\sqrt{3}) \times 193}$

$:\implies \bf{a^{3} + b^{3} = (7 + 7) \times 193}$

$:\implies \bf{a^{3} + b^{3} = 14 \times 193}$

$:\implies \bf{a^{3} + b^{3} = 2702}$

$\therefore \purple{\bf{a^{3} + b^{3} = 2702}}$

Thus the value of x³ + y³ is 2702.

Answer 2

$\bf \orange{Question : - }$

If x = (1)/(7+4√3) & y = (1)/(7-4√3) then find the value of x³ + y³ ?

$\mathsf \green{ANSWER }$

Given : -

x = (1)/(7+4√3) & y = (1)/(7-4√3)

Required to find : -

value of x³ + y³

Formula used : -

x³ + y³ = (x+y) (x²-xy+y²)

Solution : -

x = (1)/(7+4√3) & y = (1)/(7-4√3)

We need to find the value of x³ + y³ ?

So,

Value of x = (1)/(7+4√3)

Here,

Let's rationalize the denominator .

Rationalising factor of 7+4√3 = 7-4√3

Multiplying the numerator & denominator with the rationalising factor .

➣ (1)/(7+4√3) x (7-4√3)/(7-4√3)

➣ (7-4√3)/([7+4√3] [7-4√3])

➣ (7-4√3)/([7]²-[4√3]²)

➣ (7-4√3)/(49-16 x 3)

➣ (7-4√3)/(49-48)

➣ (7-4√3)/(1)

➣ (7-4√3)

Similarly,

Value of y = (1)/(7-4√3)

Here,

Let's rationalize the denominator.

Rationalising factor of 7-4√3 = 7+4√3

Multiplying the numerator & denominator with the rationalising factor .

$\sf \pink{(1)/(7-4√3) x (7+4√3)/(7+4√3)}$

$\rm{(7+4√3)/([7-4√3] [7+4√3])}$

$\sf(7+4√3)/([7]²-[4√3]²)$

$\sf(7+4√3)/(49-48)$

$\sf(7+4√3)/(1)$

$\sf(7+4√3)$

Hence,

Value of x = 7-4√3

Value of y = 7+4√3

For the next calculations let's us these values only .

Now,

Let's find the value of x² & y²

This implies;

x² = (7-4√3)²

This is in the form of;

(x-y)² = x²+y²-2xy

✯ (7)²+(4√3)²-2(7)(4√3)

✯ 49+16 x 3-48√3

✯ 49+48-48√3

✯ 97-48√3

Similarly,

✯ y² = (7+4√3)²

✯ (7)²+(4√3)²+2(7)(4√3)

✯ 49+16 x 3+48√3

✯ 49+48+48√3

✯ 97+48√3

So,

Value of x² = 97-48√3

Value of y² = 97+48√3

Now,

Let's find the value of x³+y³

Using the formula;

x³ + y³ = (x+y) (x²-xy+y²)

This implies;

$\sf{(7-4√3+[7+4√3]) ([97-48√3] -[7-4√3][7+4√3]+[97+48√3])}$

$\sf{(7-4√3+7+4√3)([97-48√3]-[{7}²-{4√3}²]+[97+58√3])}$

$\sf{(7+7)(97-48√3-[49-48]+97+58√3)}$

$\sf{(14)(97-48√3-1+97+58√3)}$

$\sf{(14)(97+97-1)}$

$\sf{(14)(194-1)}$

$\sf{(14)(193)}$

$\sf \huge \red{➨ \: 2702 }$

Therefore,

$\rm{Value \: of \: x³+y³ = \pink{ 2,702}}$