Question

find a b + BC + CA if a + b + c = 29 and a² + b² + c² = 35

Answer 1

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✴✴{Here is your answer↓}⬇⏬⤵

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▶⏩ It is given that:-)

↪➡{a + b + c = 29.}
​ ↪➡ {a² + b² + c² = 35}.

▶⏩ Now, 

↪➡a+b+c=29. 

{squaring both side }

↪➡(a² + b² +  c²) = (29)².

▶⏩using identity :-) 

▶⏩(a+b+c)²=a²+b²+c²+2(ab+bc+ca).​

↪➡a²+b²+c²+2(ab+bc+ca)=841.

↪➡35+2(ab+bc+ca)=841. 

↪➡2(ab+bc+ca)=841−35. 

↪➡2(ab+bc+ca)=846. 

↪➡{ ab + bc + ca }= {846}/{2}.

↪➡ab+bc+ca=403.

✅✅ Hence, it is finded. ✔✔

✴✴{THANKS} ✴✴

☺☺☺{Hope it is helpful for you}✌✌✌.

Answer 2

=> a + b + c = 29. ____[◢GIVEN]

=> a² + b² + c² = 35. _____[◢GIVEN]

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=> a + b + c = 29

SQUARING BOTH SIDE :-

=> (a + b + c)² = (29)² _________________[◢Eq(1)]

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●USING IDENTITY :-

$= > {( \alpha + \beta + \gamma )}^{2} = { \alpha }^{2} + { \beta }^{2} + { \gamma }^{2} + 2( \alpha \beta + \beta \gamma + \gamma \alpha )$
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●BY Eq(1)

=> a²+b²+c²+2(ab+bc+ca) = 841

=> 35 + 2(ab + bc + cd) = 841

=> 2(ab + bc + cd) = 841 - 35 = 806

=> ab + bc + cd = 403. __[◢ANSWER]

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