Question

If a+b+c = 10 , ab+bc+ca =10 then find a²+b²+c²​

Answer 1

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{❖HOLA!}}}$

$\large\orange{\sf{Given÷}}$

$\large\blue{\sf{i.~a+b+c~=~10}}$

$\large\blue{\sf{ii.~ab+bc+ca~=~10}}$

$\large\orange{\sf{To~find~÷}}$

$\large\blue{\sf{a²+b²+c²}}$

$\large\orange{\sf{Solution÷}}$

$\fbox\pink{Using~identity~(a+b+c)²~=~a²+b²+c²+ab+bc+ca}$

$\large\blue{\sf{Putting~the~values}}$

$\large\blue{\sf{(10)²~=~a²+b²+c²+10}}$

$\large\blue{\sf{100~=~a²+b²+c²+10}}$

$\large\blue{\sf{a²+b²+c²~=~90}}$

$\green{✤✤-Hope~it~helps-✤✤}$

Answer 2

EXPLANATION.

⇒ a + b + c = 10.

⇒ ab + bc + ca = 10.

As we know that,

Formula of :

⇒ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Using this formula in the equation, we get.

⇒ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Put the values in the equation, we get.

⇒ (10)² = a² + b² + c² + 2(ab + bc + ca).

⇒ (10)² = a² + b² + c² + 2(10).

⇒ 100 = a² + b² + c² + 20.

⇒ 100 - 20 = a² + b² + c².

⇒ 80 = a² + b² + c².

⇒ a² + b² + c² = 80.

                                                                                                                       

MORE INFORMATION.

(1) = (x + y)² = x² + y² + 2xy.

(2) = (x - y)² = x² + y² - 2xy.

(3) = (x² - y²) = (x + y)(x - y).

(4) = (x² + y²) = (x + y)² - 2xy.

(5) = (x + y)³ = x³ + 3x²y + 3xy² + y³.

(6) = (x - y)³ = x³ - 3x²y + 3xy² - y³.

(7) = (x³ + y³) = (x + y)(x² - xy + y²).

(8) = (x³ - y³) = (x - y)(x² + xy + y²).