Question

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Answer 1

$\bf\large\underline\blue{Question:-}$

• If $2x+3y=13$ and $xy=6$, find the value of $8x^{3}+27{y}^{3}$.

$\bf\large\underline\blue{Solution:-}$

• Before we start, we have to know the identity to be used while solving the problem.
• Here is the identity$\downarrow$

${x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)$

Now, we will solve the problem.

$2x + 3y = 13$

$xy = 6$

$\bf\underline\blue{Therefore,}$

$8 {x}^{3} + 27 {y}^{3}$

$= {(2x)}^{3} + {(3y)}^{3}$

$= {(2x + 3y)}^{3} - 2 \times 2x \times 3y(2x + 3y)$

$= {(2x + 3y)}^{3} - 12xy(2x + 3y)$

$= {13}^{3} - 12 \times 6 \times 13$

$= 2197 - 936$

$= 1261$

$\bf\large\underline\blue{Answer:-}$

• $8 {x}^{3} + 27 {y}^{3} = 1261$

Answer 2

◕ Question :

If 2x + 3y = 13 and xy = 6 then find the value of 8x³ + 27y³.

◕ To Find :

The value of 8x³ and 27y³.

◕ Given :

• (x + y) = 13
• xy = 6

◕ We Know :

• $\sf{(a + b)^{3} = (a + b)(a^{2} - ab + b^{2})}$

• $\sf{(a + b)^{2} + 2ab = a^{2} + b^{2}}$

◕ Solution :

$\purple{\sf{8x^{3} + 27y^{3}}} \\ \\ \\ \implies \sf{(2x)^{3} + (3y)^{3}}$

Using the Identity ,

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

We get :

$\implies \sf{(2x + 3y)\big((2x)^{2} - 6xy + (3y)^{2}\big)}$

Putting the value of (x + y) and (xy) , we get :

$\implies \sf{(13)\big((2x)^{2} - 6 \times 6 + (3y)^{2}\big)} \\ \\ \\\implies \sf{(13)\big((2x)^{2} - 36 + (3y)^{2}\big)}$

Using the identity :

$\sf{(a + b)^{2} - 2ab = a^{2} + b^{2}}$

We Get :

$\implies \sf{(13)\big((2x + 3y)^{2} - 2ab - 36\big)}$

Putting the value of (x + y) and (xy) , we get :

$\implies \sf{(13)\big(13^{2} - 12 \times 6 - 6\big)}\\ \\ \\ \implies \sf{(13)\big(169 - 72 - 36\big)} \\ \\ \\ \implies \sf{(13)\big(169 - 108\big)} \\ \\ \\ \implies \sf{13 \times 61} \\ \\ \\ \implies \sf{793} \\ \\ \\ \therefore \purple{\sf{8a^{3} + 27b^{3} = 793}}$