Question

if ab=x and 1/a^2 +1/b^2 =y then find the value of (a+b) ^4

Answer 1

[a²+b²]/a²b² = y

[a²+b²] = y(ab)²
a²+b² = yx²

so

(a+b)⁴
= (a²+b²+2ab)²
= (yx² + 2x)²
= x²[yx+2]²
= y²x⁴ + 4x² + 4yx³

answer is
x²[y²x² + 4 + 4xy]


hope it helps you
@di

Answer 2

Concept:

From the algebraic identities, we know that the square of sum of two numbers is given by the sum of the sum of squares of that two numbers and twice of their product.

$(a+b)^2=a^2+b^2+2ab$

Given:

Given that the values are ab=x, $\frac{1}{a^2}+\frac{1}{b^2}=y$

Find:

The value of the algebraic expression $(a+b)^4$.

Solution:

Given that, ab = x

Calculating the given relation,

$\frac{1}{a^2}+\frac{1}{b^2}=y$

$\frac{a^2+b^2}{a^2b^2}=y$

$a^2+b^2=ya^2b^2$, multiplying $a^2b^2$ on both sides

$a^2+b^2=y.(ab)^2$

$a^2+b^2=x^2y$, substituting the value of ab = x

Now calculating the required

$(a+b)^4$

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

$=(a^2+b^2+2ab)^2$, applying the algebraic identities

$=(x^2y+2x)^2$, substituting the values

$=(x^2y)^2+(2x)^2+2.x^2y.2x$, applying the algebraic identities

$=x^4y^2+4x^2+4x^3y$

Hence the value of $\mathbf{(a+b)^4=x^4y^2+4x^2+4x^3y}$.

#SPJ2