Question

6. If a + b + c = 9 and ab + bc + ca=26, find a + b2 + c?.​

Answer 1

Answer:

Given, a+b+c=9 and ab+bc+ca=26.

We know, (a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca

⇒(a+b+c)

2

=a

2

+b

2

+c

2

+2(ab+bc+ca).

Putting the values here, we get,

(9)

2

=a

2

+b

2

+c

2

+2(26)

⇒81=a

2

+b

2

+c

2

+52

⇒a

2

+b

2

+c

2

=81−52=29.

Answer 2

★ Correct Question :-

If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c².

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$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question}}}$

Here the Concept of Algebraic Identity has been used. We see we are given two expressions and we need to find the value of third one. Firstly we can apply the identity and make relation between the given terms. After doing that we can equate the values and find the answer.

Let's do it !!

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★ Identity Used :-

$\\\;\boxed{\sf{\pink{(x\:+\:y\:+\:z)^{2}\;=\;\bf{x^{3}\:+\:y^{3}\:+\:z^{3}\:+\:2xy\:+\:2yz\:+\:2xz}}}}$

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★ Solution :-

Given,

» a + b + c = 9

» ab + bc + ac = 26

Let's firstly use identity to simplify.

We know that,

$\\\;\sf{\rightarrow\;\;(x\:+\:y\:+\:z)^{2}\;=\;\bf{x^{3}\:+\:y^{3}\:+\:z^{3}\:+\:2xy\:+\:2yz\:+\:2xz}}$

• Here x = a

• Here y = b

• Here z = c

By applying this we get,

$\\\;\sf{\Longrightarrow\;\;(a\:+\:b\:+\:c)^{2}\;=\;\bf{\green{a^{3}\:+\:b^{3}\:+\:c^{3}\:+\:2ab\:+\:2bc\:+\:2ac}}}$

Now grouping the terms, we get

$\\\;\sf{\Longrightarrow\;\;(a\:+\:b\:+\:c)^{2}\;=\;\bf{\green{a^{3}\:+\:b^{3}\:+\:c^{3}\:+\:2(ab\:+\:bc\:+\:ac)}}}$

We already have some values. Let's apply them now.

$\\\;\sf{\Longrightarrow\;\;(9)^{2}\;=\;\bf{a^{3}\:+\:b^{3}\:+\:c^{3}\:+\:2(26)}}$

$\\\;\sf{\Longrightarrow\;\;81\;=\;\bf{a^{3}\:+\:b^{3}\:+\:c^{3}\:+\:52}}$

$\\\;\sf{\Longrightarrow\;\;a^{3}\:+\:b^{3}\:+\:c^{3}\:+\:52\;=\;\bf{81}}$

$\\\;\sf{\Longrightarrow\;\;a^{3}\:+\:b^{3}\:+\:c^{3}\;=\;\bf{81\;-\;52}}$

$\\\;\sf{\Longrightarrow\;\;a^{3}\:+\:b^{3}\:+\:c^{3}\;=\;\bf{\red{29}}}$

This is our answer.

$\\\;\underline{\boxed{\tt{a^{2}\:+\:b^{2}\:+\:c^{2}\;=\;\bf{\purple{29}}}}}$

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★ More Identities to know :-

$\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\;+\;b^{2}\;-\;2ab}$

$\\\;\sf{\leadsto\;\;(a\:-\:b)(a\:+\:b)\;=\;a^{2}\:-\:b^{2}}$

$\\\;\sf{\leadsto\;\;(a\:+\:b)^{2}\:=\;a^{2}\:+\:b^{2}\:+\:2ab}$

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

$\\\:\sf{\leadsto\:\:(x\:+\:a)(x\:+\:b)\;=\;x^{2}\:+\:(a\:+\:b)x\:+\:ab}$