Question

If $x^a^3*x^b^3*x^(3ab(a+b))=(2^5)^(25)$ and $a+b=5$, then the value of x

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

if x^(a³) * x^(b³) * x^{3ab(a+b)} = (2^5)^(25) and (a + b) = 5 , find x ?

Sᴏʟᴜᴛɪᴏɴ :-

→ x^(a³) * x^(b³) * x^{3ab(a+b)} = (2^5)^(25)

using a^b * a^c * a^d = a^(b + c + d) in LHS we get,

→ x^(a³ + b³ + 3ab(a + b)) = (2^5)^(25)

Now, we know that, (a³ + b³ + 3ab(a + b)) = (a + b)³

So,

→ x^(a + b)³ = (2^5)^(25)

Now using (a^b)^c = (a)^(b*c) in RHS,

→ x^(a + b)³ = (2)^(5 * 25)

→ x^(a + b)³ = (2)^125

→ x^(a + b)³ = (2)^(5³)

Putting (a + b) = 5 in LHS Now,

→ x^(5³) = (2)^(5³)

comparing Now, we get,

→ x = 2 (Ans.)

Answer 2

${ \boxed{ \bold{ \boxed{ \red{ Question:-}}}}}$

▪ If

${ \bold{ {x}^{ {a}^{3} } \times {x}^{{b}^{3} } \times {x}^{(3a(a + b))} =(( {2})^{ {5})^{25} } }}$

and ( a + b ) = 5

then , find the value of x

${ \boxed{ \bold{ \boxed{ \red{Solution:-}}}}}$

${ \bold{ \underline{ \orange{Given-}}}}$

${ \bold{ \purple{ {x}^{ {a}^{3} } \times {x}^{ {b}^{3} } \times {x}^{(3ab(a + b))} =( ({2})^{ {5})^{25} } }}}$

▪ first solving the L.H.S......

${ \bold{ = {x}^{ {a}^{3} } \times {x}^{ {b}^{3} } \times {x}^{(3ab(a + b))} }}$

▪ we know that.....

${ \boxed{ \red{ \bold{ {a}^{ \alpha } \times {a}^{ \beta } \times {a}^{ \gamma } = {a}^{( \alpha + \beta + \gamma) }}}}}$

▪ using this property in L.H.S., we get...

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

${ \boxed{ \red{ \bold{ {a}^{3} + {b}^{3} + 3ab(a + b) = {(a + b)}^{3} }}}}$

▪ now, using this algebraic identity in L.H.S...

${ \bold{ = {x}^{ {(a + b)}^{3} } }}$

▪ solving the R.H.S....

${ \bold{ = (({2})^{ {5})^{25} } }}$

${ \bold{ = {2}^{5 \times 25} = {2}^{125} }}$

▪ we know that 125 is the cube root of 5...

${ \bold{ = {2}^{ {5}^{3} }}}$

${ \bold{ \implies{ \purple{ {x}^{ {(a + b)}^{3} } = {2}^{ {5}^{3} }}}}}$

▪ In the question, it's given that..

( a + b ) = 5

▪ now, substituting the value of (a+b) in L.H.S..

${ \bold{ \implies{ {x}^{ {5}^{3} } = {2}^{ {5}^{3}}}} }$

▪ on comparing, both sides.....

${ \huge{ \boxed{ \bold{ \implies{ \red{x = 2}}}}}}$