Question

$\huge{\fbox{\fbox{\orange{\mathfrak{Explanation\:required}}}}}$​

Answer 1

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$\large{\fbox{\fbox{\orange{\mathbb{\bold{ANSWER\:1.}}}}}}$

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Answer 2

Given,

• $\sf{x^{2} - 2ax + a^{2} = 0}$

To Find,

• $\sf{The \ Value \ of \ \dfrac{x}{a}}$

Solution :

║In this Question the We use the Algebraic Identities of Square ║

→Use Identity - (a - b)² = a² + b² - 2ab

Therefore,

$\implies \sf{x^{2} - 2ax + a^{2} =0}$

$\implies \sf{(x - a)^{2} = 0}$

$\implies \sf{x - a = \sqrt{0} }$

$\implies \sf{x - a = 0}$

$\implies \sf{x = a}$

$\implies \boxed{\sf{\dfrac{x}{a} = 1}}$

Here Value of x/a = 1 Because If the value (x = a) upper side also prove then a/a = 1 or x/x = 1 So x/a = 1

$\rule{200}2$

$\underline{\boxed{\bold{\pink{Other \ Identities:}}}}$

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

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

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

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

$\implies \sf{ a^{3} + b^{3} + c^{3} -3abc = (a+b+c)(a^{2} + b^{2}+ c^{2}- ab - bc - ca)}$

$\rule{200}1$