Question

In a= √15 +4√14 and b= √15 -4 √14 find a^3 +b^3​

Answer 1

Answer :-

Given that :-

$\sf{a = \sqrt{15} + 4\sqrt{14}}$

$\sf{b = \sqrt{15} - 4\sqrt{14}}$

Now as we have this Identity :-

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

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

Now by putting the values of "a" and "b"

$\implies \tiny{\sf{a^3 + b^3 = (\sqrt{15} + 4\sqrt{14} + \sqrt{15} - 4\sqrt{14})^3 - 3(\sqrt{15} + 4\sqrt{14})(\sqrt{15} - 4\sqrt{14})(\sqrt{15} +4\sqrt{14}+\sqrt{15} - 4\sqrt{14})}}$

$\implies \sf{a^3 + b^3 = (2\sqrt{15})^3 - 3(\sqrt{15}^2 - (4\sqrt{14})^2)(2\sqrt{15})}$

$\implies \sf{ a^3 + b^3 = 8(15\sqrt{15}) - 3(15 - 224)(2\sqrt{15})}$

$\implies \sf{a^3 + b^3 = 120\sqrt{15} -3(-209)(2\sqrt{15})}$

$\implies \sf{a^3 + b^3 = 120\sqrt{15} + 1254\sqrt{15}}$

$\implies \sf{a^3 + b^3 = 1374\sqrt{15}}$

Answer 2

Answer:

$\large{\underline{\boxed{\sf a {}^{3} + {b}^{3 } = 1374 \sqrt{15}}}}$

Step-by-step explanation:

Given that :

$\sf a = \sqrt{15} + 4 \sqrt{14} \\ \sf b = \sqrt{15} - 4 \sqrt{14}$

Now, find (a + b) -

$\sf a + b = \sqrt{15} + 4 \sqrt{14} + \sqrt{15} - 4 \sqrt{14} \\ \\ \sf a + b = 2 \sqrt{15}$

Also, find the product of ab-

$\sf ab = ( \sqrt{15} + 4 \sqrt{14})( \sqrt{15} - 4 \sqrt{14} ) \\ \\ \sf ab = ( \sqrt{15}) {}^{2} - (4 \sqrt{14} )^{2} \\ \\ \sf ab = 15 - 224 \\ \\ \sf ab = - 209$

We will use identity -

• (a + b)³ = a³ + b³ + 3ab(a + b)

$\sf \implies a + b = 2 \sqrt{15} \\ \\ \bf cubing \: both \: sides - \\ \\ \sf \implies (a + b) {}^{3} = (2 \sqrt{15}) {}^{3} \\ \\ \implies \sf a {}^{3} + {b}^{3 } + 3ab(a + b) = 120\sqrt{15} \\ \\\implies \sf a {}^{3} + {b}^{3 } + 3 ( - 209)(2 \sqrt{15} ) = 120 \sqrt{15} \\ \\ \implies \sf a {}^{3} + {b}^{3 } - 627(2 \sqrt{15} ) = 120 \sqrt{15} \\ \\ \implies \sf a {}^{3} + {b}^{3 } - 1254 \sqrt{15} = 120 \sqrt{15} \\ \\ \implies \sf a {}^{3} + {b}^{3 } = 120 \sqrt{15} + 1254 \sqrt{15} \\ \\ \boxed{\bf a {}^{3} + {b}^{3 } = 1374 \sqrt{15} }$