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Using the identity \( (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \), find the value of \( (x+y+z)^2 \) if \(x=1\), \(y=2\), \(z=3\).पहचान \( (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \) का उपयोग करके \( (x+y+z)^2 \) का मान निकालिए, यदि \(x=1\), \(y=2\), \(z=3\) हों।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(36\)\(36\)
We use \( (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) \). Substitute \(x=1\), \(y=2\), \(z=3\): \(x^2+y^2+z^2=1^2+2^2+3^2=1+4+9=14\). Also \(xy+yz+zx=1\cdot2+2\cdot3+3\cdot1=2+6+3=11\). Therefore \((x+y+z)^2=14+2\cdot11=14+22=36\).हम पहचान का उपयोग करते हैं: \( (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) \). \(x=1\), \(y=2\), \(z=3\) रखने पर \(x^2+y^2+z^2=1^2+2^2+3^2=1+4+9=14\). और \(xy+yz+zx=1\cdot2+2\cdot3+3\cdot1=2+6+3=11\). अतः \((x+y+z)^2=14+2\cdot11=14+22=36\).
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